Original questions in the style of Papers 1, 2 and 3, each with a full worked mark scheme. Tick questions to build a printable test. Topic 4 at Standard Level →
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Original questions in the style of each exam boardFull worked mark schemes with method and accuracy marksCreated by Evan Kotronis with AI assistanceUpdated 14 October 2026 · new questions added continuouslyFree access until 30 November 2026How every question is checked →
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1AAHL4-q00024.1 Sampling, bias and outliersPaper 1Easy6 marks
A college has 750 students, numbered 001 to 750. A simple random sample of 5 students is chosen using the random three-digit numbers below, read from left to right. A number is used only if it corresponds to a student who has not already been selected.
512 093 781 512 000 347 906 228 650 119
(a)
Write down the numbers of the five students selected.
(2)
(b)
State what makes this a simple random sample.
(1)
(c)
The numbers of hours that the five selected students spent on voluntary work last term were 12, 7, 15, 0 and 9. Use the sample to estimate the total number of hours of voluntary work done by all 750 students last term.
(2)
(d)
Suggest one reason why this estimate may be unreliable.
(1)
Show mark scheme
Working
Mark
Notes
(a)
rejecting 781, 000 and 906 (no such student)
A1
rejecting the repeated 512: students 512, 093, 347, 228, 650
A1
(b)
every possible group of 5 students has the same probability of being chosen
R1
Accept “each student is equally likely to be chosen and choices are made at random without repetition”.
(c)
sample mean \(=\dfrac{43}{5}=8.6\) hours
(M1)
\(750\times8.6=6450\) hours
A1
(d)
the sample is very small (5 out of 750), so the sample mean may differ greatly from the population mean
R1
Accept other valid reasons, e.g. self-reported hours may be inaccurate.
2AAHL4-q00114.3 Measures of central tendency and dispersionPaper 1Easy16 marks
The marks, \(x\), of two classes in the same test are summarized below. For each class the standard deviation is that of the marks of all the students in the class.
Class
Number of students
Mean mark
Standard deviation
P
10
60
9
Q
8
69
6
For a set of \(n\) marks, the variance is \(\sigma^{2}=\dfrac{\sum x^{2}}{n}-\bar{x}^{2}\).
(a)
Show that, for class P, \(\sum x^{2}=36\,810\).
(2)
(b)
For class Q, find \(\sum x\) and \(\sum x^{2}\).
(2)
(c)
Find the mean mark of all 18 students.
(2)
(d)
Show that the standard deviation of the marks of all 18 students is 9.
(3)
(e)
The teacher scales every mark using \(y=1.2x-4\). Write down the mean and the standard deviation of the scaled marks of all 18 students.
(2)
(f)
Before scaling, it is discovered that one mark in class Q was recorded as 61 instead of 69. Find the correct mean and the correct standard deviation of the marks of class Q.
(3)
(g)
For class P, the lower quartile is 54 and the upper quartile is 62. The lowest mark in class P is 42. Determine whether this mark is an outlier.
3AAHL4-q00124.4 Correlation and regression of y on xPaper 1Easy8 marks
Different masses, \(x\) kg, were hung from a spring and its length, \(y\) cm, was measured. One length was not recorded.
\(x\) (kg)
2
4
5
7
8
10
\(y\) (cm)
13
18
22
\(w\)
31
37
The regression line of \(y\) on \(x\) for the complete data is \(y=3x+6.5\).
(a)
Write down the mean of the masses.
(1)
(b)
Using the fact that the regression line passes through the mean point, find the value of \(w\).
(3)
(c)
Interpret, in context, the value 3 and the value 6.5 in the equation of the line.
(2)
(d)
Use the regression line to estimate the length of the spring when a mass of 6 kg is hung from it.
(1)
(e)
Explain why the line should not be used to estimate the length when a mass of 25 kg is hung from it.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(\bar{x}=6\)
A1
(b)
\(\bar{y}=3(6)+6.5=24.5\)
M1
\(13+18+22+w+31+37=6\times24.5=147\)
M1
\(w=26\)
A1
(c)
3: the length increases by 3 cm for each extra kilogram
R1
6.5: the (estimated) length of the spring with no mass attached is 6.5 cm
R1
(d)
\(24.5\) cm
A1
(e)
25 kg is far outside the range of the data (extrapolation); the spring may be overstretched and stop behaving linearly
R1
Answers: (a) 6 · (b) \(w=26\) · (d) 24.5 cm
4AAHL4-q00164.5 Sample spaces and probabilityPaper 1Easy6 marks
The seven letters of the word PLANETS are arranged in a random order. All arrangements are equally likely.
(a)
Write down the number of possible arrangements.
(1)
(b)
Show that the probability that the two vowels, A and E, are next to each other is \(\dfrac{2}{7}\).
(3)
(c)
Find the probability that the first letter and the last letter are both consonants.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(7!=5040\)
A1
(b)
treating AE as one block: \(6!\) arrangements of 6 objects
M1
multiplying by 2 for the order AE or EA: \(2\times6!=1440\)
M1
\(\dfrac{1440}{5040}\)
A1
\(=\dfrac{2}{7}\)
AG
(c)
\(\dfrac{5}{7}\times\dfrac{4}{6}\)
(M1)
\(\dfrac{10}{21}\)
A1
Answers: (a) 5040 · (c) \(\tfrac{10}{21}\)
5AAHL4-q00224.6 Combined, conditional and independent eventsPaper 1Easy8 marks
Three students, Ali, Bo and Cy, each try to solve a puzzle, independently of each other. The probabilities that Ali, Bo and Cy solve it are \(\dfrac{1}{2}\), \(\dfrac{1}{3}\) and \(\dfrac{1}{4}\) respectively. The puzzle is solved if at least one of them solves it.
(a)
Find the probability that the puzzle is solved.
(2)
(b)
Find the probability that exactly one of the students solves the puzzle.
(2)
(c)
Given that the puzzle is solved, find the probability that exactly one of the students solved it.
(2)
(d)
A fourth student, who solves the puzzle with probability \(p\) independently of the others, joins them. Find the least value of \(p\) for which the probability that the puzzle is solved is at least 0.9.
7AAHL4-q00404.9 The normal distributionPaper 1Easy8 marks
A machine dispenses coffee into cups. The volume, \(X\) ml, of coffee in a cup is normally distributed with mean \(\mu\) ml and standard deviation \(\sigma\) ml. Approximately 16% of cups contain more than 215 ml and approximately 2.5% of cups contain less than 185 ml.
In this question use the facts that approximately 68%, 95% and 99.7% of a normal distribution lie within one, two and three standard deviations of the mean.
(a)
Write down two equations in \(\mu\) and \(\sigma\).
(2)
(b)
Hence find the value of \(\mu\) and the value of \(\sigma\).
(2)
(c)
Estimate the percentage of cups that contain between 195 ml and 225 ml.
(2)
(d)
The machine fills 2000 cups. Estimate the number of these cups that contain more than 235 ml.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mu+\sigma=215\)
A1
Since \(\tfrac{100-68}{2}=16\%\) lies above \(\mu+\sigma\).
\(\mu-2\sigma=185\)
A1
Since \(\tfrac{100-95}{2}=2.5\%\) lies below \(\mu-2\sigma\).
(b)
attempt to solve simultaneously, e.g. \(3\sigma=30\)
M1
\(\sigma=10\), \(\mu=205\)
A1
Both values required.
(c)
195 to 225 is \(\mu-\sigma\) to \(\mu+2\sigma\): \(34\%+47.5\%\)
(M1)
FT their \(\mu,\sigma\).
\(81.5\%\)
A1
(d)
\(235=\mu+3\sigma\), so proportion \(\approx\dfrac{100-99.7}{2}\%=0.15\%\)
8AAHL4-q00504.11 Conditional probability and independencePaper 1Easy8 marks
Events \(A\) and \(B\) are such that \(\mathrm{P}(A\mid B)=\dfrac{1}{3}\), \(\mathrm{P}(B\mid A)=\dfrac{1}{4}\) and \(\mathrm{P}(A\cap B)=\dfrac{1}{10}\).
(a)
Find \(\mathrm{P}(A)\) and \(\mathrm{P}(B)\).
(2)
(b)
Find \(\mathrm{P}(A\cup B)\).
(2)
(c)
Determine whether \(A\) and \(B\) are independent.
(2)
(d)
Find \(\mathrm{P}(A'\mid B')\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(B)=\dfrac{\mathrm{P}(A\cap B)}{\mathrm{P}(A\mid B)}\) or \(\mathrm{P}(A)=\dfrac{\mathrm{P}(A\cap B)}{\mathrm{P}(B\mid A)}\)
9AAHL4-q00544.12 Standardization and unknown normal parametersPaper 1Easy6 marks
The masses, \(X\) grams, of eggs laid by a breed of hen are normally distributed with mean \(\mu\) and standard deviation \(\sigma\). Let \(Z\) be the standard normal variable, \(Z\sim\mathrm{N}(0,\,1)\). It is given that \(\mathrm{P}(X\lt 52)=\mathrm{P}(Z\lt 0.8)\) and \(\mathrm{P}(X\lt 37)=\mathrm{P}(Z\lt -1.2)\).
(a)
Find the value of \(\mu\) and the value of \(\sigma\).
(4)
(b)
Given that \(\mathrm{P}(Z\lt 1)=0.8413\), find \(\mathrm{P}(X\gt 53.5)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{52-\mu}{\sigma}=0.8\) or \(\dfrac{37-\mu}{\sigma}=-1.2\)
10AAHL4-q00604.13 Bayes’ theoremPaper 1Easy6 marks
In a large population, 5% of people have a certain condition. A screening test gives a positive result for 90% of people who have the condition and for 10% of people who do not have the condition. A person is chosen at random and tested.
(a)
Find the probability that the test gives a positive result.
(2)
(b)
Given that the test gives a positive result, find the probability that the person has the condition.
(2)
(c)
Of 1000 people tested, find the expected number who test positive but do not have the condition, and hence comment on the result in part (b).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(0.05\times0.9+0.95\times0.1\)
(M1)
\(0.14\)
A1
(b)
\(\dfrac{0.045}{0.14}\)
(M1)
Bayes’ theorem or tree diagram; FT their (a).
\(\dfrac{9}{28}\)
A1
Accept \(0.321\).
(c)
\(1000\times0.95\times0.1=95\)
A1
only 45 positives are expected from people with the condition, so most positive results are false positives (because the condition is rare)
R1
Answers: (a) 0.14 · (b) \(\frac{9}{28}\) · (c) 95
11AAHL4-q00614.13 Bayes’ theoremPaper 1Easy6 marks
An insurance company classifies its drivers into three age groups. The tree diagram shows the proportion of drivers in each age group and the probability that a driver in that group makes a claim in a year.
Diagram not to scale
A driver is chosen at random.
(a)
Find the probability that the driver makes a claim in a year.
(2)
(b)
Given that the driver makes a claim, find the probability that the driver is under 25.
(2)
(c)
Given that the driver does not make a claim, find the probability that the driver is over 60.
12AAHL4-q00644.13 Bayes’ theoremPaper 1Easy7 marks
In a town, the probability that it rains on a given day is 0.3. When it rains, the evening forecast for that day had predicted rain with probability 0.8. When it does not rain, the forecast had predicted rain with probability 0.1.
(a)
Find the probability that the forecast predicts rain.
(2)
(b)
Given that the forecast predicts rain, find the probability that it rains.
(2)
(c)
Given that the forecast does not predict rain, find the probability that it does not rain.
(2)
(d)
State, with a reason, whether a forecast of rain or a forecast of no rain is more reliable.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(0.3(0.8)+0.7(0.1)\)
(M1)
\(0.31\)
A1
(b)
\(\dfrac{0.24}{0.31}\)
(M1)
\(\dfrac{24}{31}\)
A1
Accept \(0.774\).
(c)
\(\dfrac{0.7(0.9)}{1-0.31}=\dfrac{0.63}{0.69}\)
(M1)
\(\dfrac{21}{23}\)
A1
Accept \(0.913\).
(d)
a forecast of no rain, since \(\dfrac{21}{23}\gt\dfrac{24}{31}\)
13AAHL4-q00734.13 Bayes’ theoremPaper 1Easy14 marks
Ana cycles to school each day by one of three routes, \(A\), \(B\) or \(C\). She chooses the route by rolling a fair six-sided die: a score of 1, 2 or 3 means route \(A\), a score of 4 or 5 means route \(B\), and a score of 6 means route \(C\).
The probability that she is late for school is \(\dfrac{1}{10}\) on route \(A\), \(\dfrac{1}{5}\) on route \(B\) and \(\dfrac{2}{5}\) on route \(C\). Let \(L\) be the event that Ana is late.
(a)
Show that \(\mathrm{P}(L)=\dfrac{11}{60}\).
(2)
(b)
Given that Ana is late, find the probability that she used route \(C\).
(2)
(c)
Given that Ana is not late, find the probability that she used route \(A\).
(3)
(d)
Find the expected number of days on which Ana is late in a period of 30 school days.
(1)
(e)
Find the probability that Ana is late on exactly one of two particular days. Assume that days are independent.
(2)
(f)
Determine whether the events \(L\) and “Ana uses route \(C\)” are independent. Justify your answer.
(2)
(g)
Ana improves her cycling on route \(C\) so that the probability of being late on route \(C\) becomes \(x\). The probabilities for routes \(A\) and \(B\) do not change, and \(\mathrm{P}(L)\) becomes \(\dfrac{1}{6}\). Find \(x\).
19AAHL4-q00904.14 Variance, continuous random variables and linear transformationsPaper 1Easy7 marks
The continuous random variable \(X\) has probability density function \(f(x)=k\sqrt{x}\) for \(0\le x\le4\), and \(f(x)=0\) otherwise, where \(k\) is a constant.
(a)
Find the value of \(k\).
(2)
(b)
Find the median of \(X\), giving your answer in the form \(p\sqrt[3]{q}\), where \(p,q\in\mathbb{Z}^{+}\) and \(q\) is prime.
21AAHL4-q00014.1 Sampling, bias and outliersPaper 1Medium9 marks
A factory makes 600 items in a day, numbered 1 to 600 in the order they are made. A systematic sample of 30 items is taken as follows: an integer \(r\) is chosen at random from 1 to 20, and items \(r,\ r+20,\ r+40,\ \ldots,\ r+580\) are selected.
(a)
Find the probability that item 137 is selected.
(1)
(b)(i)
Find the probability that
items 137 and 157 are both selected;
(1)
(b)(ii)
items 137 and 140 are both selected.
(1)
(c)(i)
Instead, a simple random sample of 30 of the 600 items is taken.
Write down the probability that item 137 is selected.
(1)
(c)(ii)
Show that the probability that items 137 and 140 are both selected is \(\dfrac{29}{11\,980}\).
(2)
(d)
Hence explain why the systematic sample is not a simple random sample, even though every item has the same probability of being selected.
(1)
(e)(i)
Find the probability that at least one of items 137 and 140 is selected
in the systematic sample;
(1)
(e)(ii)
in the simple random sample.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(137=17+6(20)\), so item 137 is selected only when \(r=17\): probability \(\dfrac{1}{20}\)
A1
(b)(i)
\(\dfrac{1}{20}\)
A1
Both are selected exactly when \(r=17\).
(b)(ii)
\(0\)
A1
137 and 140 differ by 3, not a multiple of 20.
(c)(i)
\(\dfrac{30}{600}=\dfrac{1}{20}\)
A1
(c)(ii)
\(\dfrac{\binom{598}{28}}{\binom{600}{30}}\) or \(\dfrac{30}{600}\times\dfrac{29}{599}\)
M1
\(=\dfrac{870}{359\,400}\)
A1
\(=\dfrac{29}{11\,980}\)
AG
(d)
in a simple random sample every set of 30 items is equally likely, but in the systematic sample some pairs (e.g. 137 and 140) can never occur together, so not all samples are equally likely
R1
(e)(i)
\(\dfrac{1}{20}+\dfrac{1}{20}=\dfrac{1}{10}\)
A1
The two events are mutually exclusive (\(r=17\) or \(r=20\)).
Answers: (a) \(\tfrac{1}{20}\) · (b)(i) \(\tfrac{1}{20}\), (ii) 0 · (c)(i) \(\tfrac{1}{20}\) · (e)(i) \(\tfrac{1}{10}\), (ii) \(\tfrac{1169}{11\,980}\)
22AAHL4-q00084.3 Measures of central tendency and dispersionPaper 1Medium8 marks
A data set has mean 40, standard deviation 5, median 38 and interquartile range 6. Each value \(x\) in the data set is transformed to \(y=ax+b\), where \(a,b\in\mathbb{R}\) and \(a\neq0\). The transformed data have mean 70 and standard deviation 15.
(a)
Find the two possible pairs of values of \(a\) and \(b\).
(4)
(b)
It is also given that the median of the transformed data is greater than its mean. Find the value of \(a\) and the value of \(b\).
(2)
(c)
For the values in part (b), find the interquartile range of the transformed data, and state which quartile of the original data becomes the lower quartile of the transformed data.
(2)
Show mark scheme
Working
Mark
Notes
(a)
standard deviation of \(y\) is \(|a|\times5=15\)
M1
\(a=3\) or \(a=-3\)
A1
\(40a+b=70\)
M1
\(a=3,\ b=-50\) and \(a=-3,\ b=190\)
A1
(b)
medians: \(3(38)-50=64\) and \(-3(38)+190=76\)
M1
\(76\gt 70\), so \(a=-3,\ b=190\)
A1
(c)
IQR \(=|-3|\times6=18\)
A1
the upper quartile of \(x\) (multiplying by a negative number reverses the order)
A1
Answers: (a) \((3,-50)\) or \((-3,190)\) · (b) \(a=-3,\ b=190\) · (c) 18; the upper quartile
23AAHL4-q00094.3 Measures of central tendency and dispersionPaper 1Medium9 marks
The table shows the number of siblings, \(x\), of each student in a group, where \(k\in\mathbb{N}\).
Number of siblings, \(x\)
0
1
2
3
4
Frequency
6
9
\(k\)
7
5
(a)
Find the set of values of \(k\) for which the median number of siblings is 2.
(3)
(b)
Given also that 1 is the only mode, find the possible values of \(k\).
(2)
(c)
Show that the mean number of siblings is less than 2 for every value of \(k\).
(2)
(d)
Find the least value of \(k\) for which the median is 2 and the mean differs from the median by less than 0.1.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(6+9=15\) values are less than 2; total \(27+k\)
M1
the median is 2 exactly when the middle value(s) lie in positions 16 to \(15+k\) (e.g. \(k=3\): 30 values, 15th and 16th values 1 and 2, median 1.5; \(k=4\): 31 values, 16th value 2)
M1
\(k\ge4\)
A1
Accept \(k\in\{4,5,6,\ldots\}\).
(b)
need \(k\lt 9\) (\(k=9\) gives two modes)
M1
\(k=4,5,6,7,8\)
A1
(c)
mean \(=\dfrac{9+2k+21+20}{27+k}=\dfrac{50+2k}{27+k}\)
M1
\(\dfrac{50+2k}{27+k}\lt 2\iff 50+2k\lt 54+2k\), which is always true
R1
(d)
\(2-\dfrac{50+2k}{27+k}=\dfrac{4}{27+k}\lt 0.1\)
M1
\(27+k\gt 40\), so \(k=14\)
A1
Check: \(k=13\) gives a difference of exactly 0.1.
25AAHL4-q00204.6 Combined, conditional and independent eventsPaper 1Medium8 marks
A bag contains \(n\) red counters and 3 blue counters, where \(n\ge1\). Two counters are taken at random from the bag without replacement.
(a)
Show that the probability that the two counters are the same colour is \(\dfrac{n^{2}-n+6}{(n+3)(n+2)}\).
(3)
(b)
Given that this probability is \(\dfrac{1}{2}\), find the possible values of \(n\).
(3)
(c)
Using the larger value of \(n\), three counters are taken at random from the bag without replacement. Find the probability that all three are the same colour.
or \(\dfrac{\binom{6}{3}+\binom{3}{3}}{\binom{9}{3}}\)
\(\dfrac{21}{84}=\dfrac{1}{4}\)
A1
Answers: (b) \(n=1\) or \(n=6\) · (c) \(\tfrac{1}{4}\)
26AAHL4-q00264.7 Discrete random variablesPaper 1Medium10 marks
The probability distribution of a discrete random variable \(X\) is given by \(\mathrm{P}(X=x)=a+bx\), for \(x\in\{1,\,2,\,3,\,4\}\), where \(a\) and \(b\) are constants.
(a)
Show that \(\mathrm{E}(X)=2.5+5b\).
(3)
(b)
By considering the conditions on a probability distribution, find the set of possible values of \(b\).
(3)
(c)
Hence write down the least possible value and the greatest possible value of \(\mathrm{E}(X)\).
(1)
(d)
Given that \(\mathrm{E}(X)=3\), find \(\mathrm{P}(X=4\mid X\ge2)\).
27AAHL4-q00274.7 Discrete random variablesPaper 1Medium9 marks
A board is divided into an \(n\times n\) grid of squares, where \(n\ge2\). A counter is placed on a square chosen at random. Let \(X\) be the number of squares that share an edge with the square on which the counter is placed.
(a)
For a \(4\times4\) grid, find the probability distribution of \(X\).
(3)
(b)
For a \(4\times4\) grid, find \(\mathrm{E}(X)\).
(1)
(c)
Show that, for an \(n\times n\) grid, \(\mathrm{E}(X)=\dfrac{4(n-1)}{n}\).
(3)
(d)
A second counter is then placed on a different square, chosen at random. Hence, or otherwise, show that the probability that the two counters are on squares that share an edge is \(\dfrac{4}{n(n+1)}\).
29AAHL4-q00314.7 Discrete random variablesPaper 1Medium15 marks
A fair coin is tossed four times. Let \(X\) be the length of the longest run of consecutive heads; for example, the outcome HHTH gives \(X=2\), and TTTT gives \(X=0\). The table shows part of the probability distribution of \(X\).
\(x\)
0
1
2
3
4
\(\mathrm{P}(X=x)\)
\(\tfrac{1}{16}\)
\(\tfrac{1}{16}\)
(a)
Show that \(\mathrm{P}(X=3)=\dfrac{1}{8}\).
(2)
(b)
Complete the table of the probability distribution of \(X\) shown above.
(3)
(c)
Show that \(\mathrm{E}(X)=\dfrac{27}{16}\).
(2)
(d)
Find \(\mathrm{Var}(X)\).
(3)
(e)
In a game, a player pays \(c\) euros to play and receives \(4X\) euros. Find the value of \(c\) for which the game is fair.
(2)
(f)
Find the variance of the amount received.
(1)
(g)
Given that \(X\ge2\), find the probability that the first toss was a head.
(2)
Show mark scheme
Working
Mark
Notes
(a)
outcomes HHHT and THHH only
M1
Both outcomes listed, and no others.
\(\dfrac{2}{16}\)
A1
\(=\dfrac{1}{8}\)
AG
(b)
listing outcomes with no two consecutive heads: TTTH, TTHT, THTT, HTTT, HTHT, THTH, HTTH
M1
Or listing the five outcomes with X = 2: HHTT, THHT, TTHH, HHTH, HTHH.
30AAHL4-q00334.8 Binomial distributionPaper 1Medium9 marks
A random variable \(X\) follows a binomial distribution \(\mathrm{B}(n,\,p)\), where \(0\lt p\lt 1\). It is given that \(\mathrm{P}(X=1)=3\,\mathrm{P}(X=0)\) and \(4\,\mathrm{P}(X=2)=5\,\mathrm{P}(X=1)\).
(a)
Show that \(np=3(1-p)\).
(2)
(b)
Show that \((n-1)p=\dfrac{5}{2}(1-p)\).
(2)
(c)
Hence find the value of \(p\) and the value of \(n\).
(3)
(d)
Find \(\mathrm{P}(X\le1)\), giving your answer as a fraction.
31AAHL4-q00484.11 Conditional probability and independencePaper 1Medium8 marks
A card is drawn at random from a standard pack of 52 playing cards (13 cards in each of the four suits hearts, diamonds, clubs and spades; each suit contains one king, one queen and one jack). Let \(H\) be the event “the card is a heart”, \(K\) the event “the card is a king” and \(C\) the event “the card is a king, a queen or a jack”.
(a)
Show that \(H\) and \(K\) are independent.
(2)
(b)
By finding \(\mathrm{P}(K\mid C)\), show that \(K\) and \(C\) are not independent.
(2)
(c)
Events \(A\) and \(B\) are mutually exclusive, with \(\mathrm{P}(A)\gt 0\) and \(\mathrm{P}(B)\gt 0\). Prove that \(A\) and \(B\) are not independent.
(2)
(d)
The two of clubs is removed from the pack and a card is drawn at random from the remaining 51 cards. Determine whether \(H\) and \(K\) are now independent.
\(\dfrac{13}{51}\times\dfrac{4}{51}=\dfrac{52}{2601}\neq\dfrac{51}{2601}=\dfrac{1}{51}\), so not independent
R1
Or \(\mathrm{P}(K\mid H)=\dfrac{1}{13}\neq\dfrac{4}{51}\).
Answers: (b) \(\mathrm{P}(K\mid C)=\frac13\) · (d) not independent
32AAHL4-q00534.11 Conditional probability and independencePaper 1Medium16 marks
In a town, if it rains on one day, the probability that it rains on the next day is \(\dfrac{3}{5}\). If it does not rain on one day, the probability that it rains on the next day is \(\dfrac{1}{5}\). It rains on day 1. Let \(p_{n}\) be the probability that it rains on day \(n\), so \(p_{1}=1\).
(a)
Find \(p_{2}\) and \(p_{3}\).
(3)
(b)
Show that \(p_{n+1}=\dfrac{1}{5}+\dfrac{2}{5}p_{n}\).
(2)
(c)
By considering \(p_{n}-\dfrac{1}{3}\), show that \(p_{n}=\dfrac{1}{3}+\dfrac{2}{3}\left(\dfrac{2}{5}\right)^{n-1}\).
(4)
(d)
Write down the value of \(p_{n}\) in the long run.
(1)
(e)
Find the least value of \(n\) for which \(p_{n}-\dfrac{1}{3}\lt\dfrac{1}{150}\).
(3)
(f)
Given that it rains on day 3, find the probability that it did not rain on day 2.
33AAHL4-q00624.13 Bayes’ theoremPaper 1Medium7 marks
A veterinary test is used on cattle in a region where 10% of cattle have a disease. The test is positive for 90% of cattle with the disease. For cattle without the disease, the test is positive with probability \(q\), where \(0\lt q\lt 1\). Let \(D\) be the event that an animal has the disease and \(+\) the event that its test is positive.
(a)
Show that \(\mathrm{P}(D\mid +)=\dfrac{1}{1+10q}\).
(2)
(b)
The test is acceptable if \(\mathrm{P}(D\mid +)\ge0.8\). Find the largest value of \(q\) for which the test is acceptable.
(2)
(c)
For this value of \(q\), find the probability that an animal whose test is negative does not have the disease.
34AAHL4-q00634.13 Bayes’ theoremPaper 1Medium10 marks
A survey estimates the proportion, \(\pi\), of drivers who have driven through a red traffic light. Each driver privately rolls a fair six-sided die. If the score is 1 or 2, the driver answers “Yes” whatever the truth; otherwise the driver answers the question “Have you ever driven through a red light?” truthfully. Let \(T\) be the event that the driver has driven through a red light and \(Y\) the event that the driver answers “Yes”.
(a)
Show that \(\mathrm{P}(Y)=\dfrac{1+2\pi}{3}\).
(2)
(b)
In the survey, 40% of drivers answer “Yes”. Find the value of \(\pi\).
(2)
(c)
Find \(\mathrm{P}(T\mid Y)\).
(2)
(d)
The procedure is changed: a score of 1 means answer “Yes”, a score of 2 means answer “No”, and otherwise the driver answers truthfully. Taking \(\pi=0.1\), find \(\mathrm{P}(T\mid Y)\) and \(\mathrm{P}(T\mid Y')\).
(4)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(Y)=\dfrac13(1)+\dfrac23\,\pi\)
M1
Two branches of the die outcome.
\(=\dfrac{1}{3}+\dfrac{2\pi}{3}\)
A1
\(=\dfrac{1+2\pi}{3}\)
AG
(b)
\(\dfrac{1+2\pi}{3}=0.4\)
M1
\(\pi=0.1\)
A1
(c)
\(\mathrm{P}(Y\mid T)=1\), so \(\mathrm{P}(T\mid Y)=\dfrac{0.1\times1}{0.4}\)
35AAHL4-q00664.13 Bayes’ theoremPaper 1Medium8 marks
In a game show a prize is placed behind one of three closed doors, 1, 2 and 3, each door being equally likely. A contestant chooses door 1. The host, who knows where the prize is, then opens one of doors 2 and 3, always choosing a door that does not hide the prize. If the prize is behind door 1, the host opens door 3 with probability \(q\) and door 2 with probability \(1-q\), where \(0\le q\le1\). The host opens door 3.
(a)
In the case \(q=\dfrac12\), show that the probability that the prize is behind door 2 is \(\dfrac23\).
(3)
(b)
For a general value of \(q\), find, in terms of \(q\), the probability that the prize is behind door 1.
(3)
(c)
The contestant may switch to door 2. Show that switching never decreases the contestant’s probability of winning.
\(\dfrac{1}{1+q}\ge\dfrac{q}{1+q}\) because \(q\le1\), so switching is never worse (equal only when \(q=1\))
R1
Answers: (b) \(\frac{q}{1+q}\) · (c) \(\mathrm{P}(\text{win by switching})=\frac{1}{1+q}\)
36AAHL4-q00684.13 Bayes’ theoremPaper 1Medium7 marks
A shop buys all its light bulbs from three suppliers, P, Q and R. Supplier P provides half of the bulbs, supplier Q provides a proportion \(x\) and supplier R provides the rest. The probability that a bulb is faulty is 0.02 for supplier P, 0.04 for supplier Q and 0.08 for supplier R. Half of all faulty bulbs come from supplier R.
(a)
Find the value of \(x\).
(4)
(b)
A bulb is chosen at random and is not faulty. Find the probability that it came from supplier P.
37AAHL4-q00694.13 Bayes’ theoremPaper 1Medium8 marks
Let \(A\) and \(B\) be events with \(0\lt\mathrm{P}(A)\lt1\) and \(0\lt\mathrm{P}(B)\lt1\).
(a)
Show that \(\mathrm{P}(B\mid A)-\mathrm{P}(B)=\dfrac{\mathrm{P}(B)\,\mathrm{P}(B')\big(\mathrm{P}(A\mid B)-\mathrm{P}(A\mid B')\big)}{\mathrm{P}(A)}\).
(4)
(b)
Hence state a condition on \(\mathrm{P}(A\mid B)\) and \(\mathrm{P}(A\mid B')\) for which knowing that \(A\) has occurred increases the probability of \(B\).
(1)
(c)
Hence show that if \(\mathrm{P}(A\mid B)=\mathrm{P}(A\mid B')\) then \(A\) and \(B\) are independent.
(1)
(d)
Given that \(\mathrm{P}(B)=0.4\), \(\mathrm{P}(A\mid B)=0.3\) and \(\mathrm{P}(A\mid B')=0.2\), use the result in part (a) to find \(\mathrm{P}(B\mid A)\).
38AAHL4-q00704.13 Bayes’ theoremPaper 1Medium9 marks
A transmitter sends the digit 1 with probability \(\dfrac35\) and the digit 0 with probability \(\dfrac25\). To reduce errors, each digit is sent three times (as 111 or 000). Each of the three received digits is independently changed (1 to 0 or 0 to 1) with probability \(\dfrac{1}{10}\).
(a)
Find the probability that the receiver gets 110, in that order.
(3)
(b)
Given that the receiver gets 110, find the probability that the digit 1 was sent.
(2)
(c)
The receiver decodes each block as the digit that appears at least twice. Find the probability that a block is decoded incorrectly.
(2)
(d)
Given that the receiver gets 100, find the probability that the digit 0 was sent.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(110\mid1)=\left(\dfrac{9}{10}\right)^{2}\dfrac{1}{10}=\dfrac{81}{1000}\) and \(\mathrm{P}(110\mid0)=\left(\dfrac{1}{10}\right)^{2}\dfrac{9}{10}=\dfrac{9}{1000}\)
Reject \(a=-\tfrac{13}{4}\) since \(a\gt1\). Award A0 if both values are given as the answer.
(c)
\(\mathrm{E}(X)=1.1\), \(\sigma=\sqrt{2.29}\) with \(1.5\lt\sigma\lt1.6\), so the interval is approximately \(]-0.4,\ 2.6[\), containing \(x=0\) and \(x=1\) only
(M1)
Since \(1.5^{2}=2.25\lt2.29\lt2.56=1.6^{2}\).
\(0.5+0.3=0.8\)
A1
Answers: (b) \(a=4\) · (c) \(0.8\)
40AAHL4-q00854.14 Variance, continuous random variables and linear transformationsPaper 1Medium8 marks
The continuous random variable \(X\) has probability density function \(f(x)=k(9-x^{2})\) for \(0\le x\le3\), and \(f(x)=0\) otherwise.
(a)
Find the value of \(k\).
(2)
(b)
Find \(\mathrm{E}(X)\).
(2)
(c)
Find \(\mathrm{P}\big(X\lt\mathrm{E}(X)\big)\).
(2)
(d)
Hence state, with a reason, whether the median of \(X\) is greater than or less than \(\mathrm{E}(X)\).
\(\mathrm{P}\big(X\lt\mathrm{E}(X)\big)=\dfrac{549}{1024}\gt\dfrac{512}{1024}=\dfrac12\), so half of the probability lies below \(\mathrm{E}(X)\) already
42AAHL4-q00884.14 Variance, continuous random variables and linear transformationsPaper 1Medium8 marks
The continuous random variable \(X\) has the probability density function \(f\) shown. The graph consists of line segments joining \((0,0)\), \((1,h)\), \((3,h)\) and \((4,0)\), and \(f(x)=0\) for \(x\lt 0\) and \(x\gt 4\).
43AAHL4-q00894.14 Variance, continuous random variables and linear transformationsPaper 1Medium9 marks
The continuous random variable \(X\) has probability density function \(f(x)=\dfrac{1}{b-a}\) for \(a\le x\le b\), and \(f(x)=0\) otherwise, where \(a\lt b\).
(a)
Show that \(\mathrm{E}(X)=\dfrac{a+b}{2}\).
(2)
(b)
Show that \(\mathrm{Var}(X)=\dfrac{(b-a)^{2}}{12}\).
(3)
(c)
Given that \(\mathrm{E}(X)=7\) and \(\mathrm{Var}(X)=3\), find the value of \(a\) and the value of \(b\).
44AAHL4-q00924.14 Variance, continuous random variables and linear transformationsPaper 1Medium8 marks
The continuous random variable \(X\) has probability density function \(f(x)=k\mathrm{e}^{-x}\) for \(0\le x\le\ln4\), and \(f(x)=0\) otherwise, where \(k\) is a constant.
(a)
Show that \(k=\dfrac43\).
(2)
(b)
Find the median of \(X\), giving your answer in the form \(\ln\dfrac{p}{q}\), where \(p,q\in\mathbb{Z}^{+}\).
(3)
(c)
Find \(\mathrm{E}(X)\), giving your answer in the form \(a+b\ln2\), where \(a,b\in\mathbb{Q}\).
47AAHL4-q00954.14 Variance, continuous random variables and linear transformationsPaper 1Medium9 marks
A machine cuts square tiles. The side length, \(X\) cm, of a randomly chosen tile is modelled by the probability density function \(f(x)=\dfrac{x-1}{2}\) for \(1\le x\le3\), and \(f(x)=0\) otherwise.
(a)
Find the expected side length.
(2)
(b)
Find the expected area of a tile.
(2)
(c)
Find the standard deviation of the perimeter of a tile.
(3)
(d)
Find the probability that the area of a randomly chosen tile is greater than \(4\ \text{cm}^{2}\).
50AAHL4-q00044.2 Presentation of dataPaper 1Hard9 marks
A data set consists of seven integers \(x_{1}\le x_{2}\le\cdots\le x_{7}\). The quartiles are found by taking the median of the lower three values and the median of the upper three values. The box-and-whisker diagram summarizes the data.
(a)
Write down the values of \(x_{1},\ x_{2},\ x_{4},\ x_{6}\) and \(x_{7}\).
(2)
(b)
Determine whether the largest value is an outlier.
(2)
(c)
Find the least possible value and the greatest possible value of the mean of the data.
(3)
(d)
Given that the mean of the data is equal to the median, find the values of \(x_{3}\) and \(x_{5}\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
recognizing \(Q_{1}=x_{2}\), median \(=x_{4}\), \(Q_{3}=x_{6}\)
only possible with the least values: \(x_{3}=6,\ x_{5}=10\)
A1
Answers: (a) 3, 6, 10, 14, 21 · (b) not an outlier · (c) 10 and \(\tfrac{78}{7}\) · (d) \(x_{3}=6,\ x_{5}=10\)
51AAHL4-q00054.2 Presentation of dataPaper 1Hard10 marks
The time, \(t\) minutes, that each of 200 visitors spent in a gallery was recorded. For \(0\le t\le60\), the cumulative frequency curve is modelled by \(F(t)=\dfrac{200}{1+\mathrm{e}^{-(t-30)/5}}\), as shown.
(a)
Write down the median time.
(1)
(b)
Show that the lower quartile is \(30-5\ln3\).
(2)
(c)
Write down the upper quartile.
(1)
(d)
Show that a visitor’s time is an upper outlier if it is greater than \(30+20\ln3\).
(2)
(e)
Use the model to estimate the number of visitors whose times are upper outliers.
(2)
(f)
Find the 90th percentile, giving your answer in the form \(a+b\ln3\), where \(a,b\in\mathbb{Z}\).
Answers: (a) 30 min · (c) \(30+5\ln3\) · (e) about 2 · (f) \(30+10\ln3\)
52AAHL4-q00174.5 Sample spaces and probabilityPaper 1Hard11 marks
Two integers \(a\) and \(b\) are chosen independently and at random from \(\{1,\,2,\,\ldots,\,n\}\), where \(n\ge2\). Every ordered pair \((a,\ b)\) is equally likely.
(a)
Write down \(\mathrm{P}(a=b)\).
(1)
(b)
Hence show that \(\mathrm{P}(a\lt b)=\dfrac{n-1}{2n}\).
(2)
(c)
Show that \(\mathrm{P}(|a-b|\le1)=\dfrac{3n-2}{n^{2}}\).
(3)
(d)
Given that \(\mathrm{P}(|a-b|\le1)=\dfrac{7}{25}\), find the value of \(n\).
(3)
(e)
For this value of \(n\), find \(\mathrm{P}(a\lt b\mid |a-b|\le1)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{n}{n^{2}}=\dfrac{1}{n}\)
A1
(b)
by symmetry \(\mathrm{P}(a\lt b)=\mathrm{P}(a\gt b)=\dfrac{1}{2}\left(1-\dfrac{1}{n}\right)\)
M1
\(=\dfrac{1}{2}\cdot\dfrac{n-1}{n}\)
A1
\(=\dfrac{n-1}{2n}\)
AG
(c)
counting pairs with \(b=a+1\): \(n-1\) pairs; similarly \(n-1\) pairs with \(a=b+1\)
M1
favourable pairs \(=n+2(n-1)=3n-2\)
A1
out of \(n^{2}\) equally likely pairs
A1
\(\dfrac{3n-2}{n^{2}}\)
AG
(d)
\(7n^{2}=25(3n-2)\Rightarrow 7n^{2}-75n+50=0\)
M1
\((n-10)(7n-5)=0\)
A1
\(n=10\) (rejecting \(n=\tfrac{5}{7}\), not an integer)
53AAHL4-q00214.6 Combined, conditional and independent eventsPaper 1Hard10 marks
In a game of tennis, when the score reaches “deuce” the game continues until one player has won two more points than the other. Ana wins each point against Bea with probability \(p\), independently of all other points, where \(0\lt p\lt 1\). The game is at deuce.
(a)(i)
Let \(p=\dfrac{2}{3}\).
Write down the probability that Ana wins the next two points.
(1)
(a)(ii)
Find the probability that the score returns to deuce after the next two points.
(2)
(b)
Hence show that, when \(p=\dfrac{2}{3}\), the probability that Ana wins the game is \(\dfrac{4}{5}\).
(3)
(c)
Show that, for general \(p\), the probability that Ana wins the game is \(\dfrac{p^{2}}{1-2p+2p^{2}}\).
(2)
(d)
Find the value of \(p\) for which the probability that Ana wins the game is 0.9.
(2)
Show mark scheme
Working
Mark
Notes
(a)(i)
\(\dfrac{4}{9}\)
A1
(a)(ii)
Ana wins exactly one of the two points: \(2\times\dfrac{2}{3}\times\dfrac{1}{3}\)
Answers: (a)(i) \(\tfrac{4}{9}\), (ii) \(\tfrac{4}{9}\) · (d) \(p=\tfrac{3}{4}\)
54AAHL4-q00254.7 Discrete random variablesPaper 1Hard10 marks
The probability distribution of a discrete random variable \(X\) is given by \(\mathrm{P}(X=x)=\dfrac{k}{x(x+1)}\), for \(x\in\{1,\,2,\,\ldots,\,9\}\), where \(k\) is a constant.
(a)
By writing \(\dfrac{1}{x(x+1)}=\dfrac{1}{x}-\dfrac{1}{x+1}\), show that \(k=\dfrac{10}{9}\).
(3)
(b)
Find \(\mathrm{P}(X\le3)\).
(2)
(c)
Find \(\mathrm{P}(X\gt 4\mid X\gt 1)\).
(3)
(d)
Find the least integer \(m\) for which \(\mathrm{P}(X\ge m)\lt 0.1\).
55AAHL4-q00294.7 Discrete random variablesPaper 1Hard15 marks
A town has \(n\) cafés, where \(n\ge3\). Each of three friends chooses one of the cafés at random, independently of the others, to visit on Saturday. Let \(X\) be the number of different cafés chosen by the three friends.
(a)(i)
In this part, \(n=4\).
Find the probability distribution of \(X\).
(3)
(a)(ii)
Hence find \(\mathrm{E}(X)\).
(1)
(b)
Show that, for general \(n\), \(\mathrm{P}(X=1)=\dfrac{1}{n^{2}}\) and \(\mathrm{P}(X=3)=\dfrac{(n-1)(n-2)}{n^{2}}\).
(3)
(c)
Hence show that \(\mathrm{P}(X=2)=\dfrac{3(n-1)}{n^{2}}\).
(2)
(d)
Show that \(\mathrm{E}(X)=3-\dfrac{3}{n}+\dfrac{1}{n^{2}}\).
(2)
(e)
Find the least value of \(n\) for which the probability that the three friends choose three different cafés is greater than 0.9.
(3)
(f)
Describe what happens to \(\mathrm{E}(X)\) as \(n\) increases, and interpret this in context.
(1)
Show mark scheme
Working
Mark
Notes
(a)(i)
e.g. \(\mathrm{P}(X=3)=1\times\dfrac{3}{4}\times\dfrac{2}{4}\)
57AAHL4-q00654.13 Bayes’ theoremPaper 1Hard9 marks
A fair six-sided die is rolled to choose a box. If the score is 1, box X is chosen; if the score is 2 or 3, box Y is chosen; otherwise box Z is chosen. Box X contains 3 red and 1 blue ball, box Y contains 2 red and 2 blue balls and box Z contains 2 red and 4 blue balls. Two balls are then taken at random from the chosen box, without replacement.
(a)
Find the probability that both balls are red.
(3)
(b)
Given that both balls are red, find the probability that box Z was chosen.
(2)
(c)
Given that exactly one of the balls is red, find the probability that box Y was chosen.
58AAHL4-q00674.13 Bayes’ theoremPaper 1Hard11 marks
A multiple-choice question has four options, exactly one of which is correct. A student knows the answer with probability \(p\), where \(0\lt p\lt 1\). If the student knows the answer, the student answers correctly; otherwise the student guesses one of the four options at random. Let \(K\) be the event that the student knows the answer and \(C\) the event that the student answers correctly.
(a)
Show that \(\mathrm{P}(C)=\dfrac{1+3p}{4}\).
(2)
(b)
Show that \(\mathrm{P}(K\mid C)=\dfrac{4p}{1+3p}\).
(2)
(c)
Find the value of \(p\) for which \(\mathrm{P}(K\mid C)=0.9\).
(2)
(d)(i)
Let \(g(p)=\mathrm{P}(K\mid C)-p\), the increase in the probability that the student knows the answer once a correct answer is seen.
Find the value of \(p\) for which \(g(p)\) is a maximum. Justify that it is a maximum.
Or completing the square, or vertex of the parabola.
\(p=\tfrac{5}{16}\), which lies in \(\left]0,\tfrac12\right]\), and \(\dfrac{\mathrm{d}^{2}}{\mathrm{d}p^{2}}=-32\lt 0\) so a maximum
A1
Must check that the value lies in the domain from (a) and justify the maximum.
maximum \(\mathrm{Var}(X)=\dfrac{50}{16}-\dfrac{25}{16}=\dfrac{25}{16}\)
A1
Answers: (a) \(0\lt p\le\tfrac12\) · (c) \(\tfrac{25}{16}\) when \(p=\tfrac5{16}\)
60AAHL4-q00824.14 Variance, continuous random variables and linear transformationsPaper 1Hard10 marks
The discrete random variable \(X\) takes only the values \(0\) and \(c\), where \(c\gt 0\), with \(\mathrm{P}(X=c)=p\) and \(\mathrm{P}(X=0)=1-p\), where \(0\lt p\lt 1\).
(a)
Show that \(\mathrm{Var}(X)=\mathrm{E}(X)\big(c-\mathrm{E}(X)\big)\).
(3)
(b)
Given that \(\mathrm{E}(X)=2\) and \(\mathrm{Var}(X)=6\), find the value of \(c\) and the value of \(p\).
(3)
(c)
For the values in part (b), the random variable \(Y=aX+b\), where \(a\gt0\), takes only the values \(-1\) and \(1\). Find the value of \(a\) and the value of \(b\), and hence find \(\mathrm{E}(Y)\) and \(\mathrm{Var}(Y)\).
(4)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{E}(X)=pc\) and \(\mathrm{E}(X^{2})=pc^{2}\)
A1
\(\mathrm{Var}(X)=pc^{2}-p^{2}c^{2}\)
M1
Use of \(\mathrm{E}(X^{2})-[\mathrm{E}(X)]^{2}\).
\(=pc\,(c-pc)\)
A1
Factorizing.
\(=\mathrm{E}(X)\big(c-\mathrm{E}(X)\big)\)
AG
(b)
\(6=2(c-2)\)
M1
Using the result of (a).
\(c=5\)
A1
\(p=\dfrac{2}{5}\)
A1
From \(pc=2\).
(c)
\(b=-1\) and \(a=\dfrac25\)
A1
From \(a(0)+b=-1\) and \(5a+b=1\).
\(\mathrm{E}(Y)=\tfrac25(2)-1\) and \(\mathrm{Var}(Y)=\left(\tfrac25\right)^{2}(6)\)
M1
Use of \(\mathrm{E}(aX+b)=a\mathrm{E}(X)+b\) and \(\mathrm{Var}(aX+b)=a^{2}\mathrm{Var}(X)\). FT their \(a\), \(b\).
Write down \(\mathrm{E}(X)\), giving a reason for your answer.
(1)
(c)
Given that \(\mathrm{Var}(X)=\dfrac{13}{45}\), find the value of \(a\) and the value of \(b\).
(4)
(d)
For these values, find \(\mathrm{P}\left(|X|\lt\dfrac{1}{2}\right)\).
(3)
(e)
For these values, write down the mode of \(X\).
(1)
(f)
Now suppose that \(a\) and \(b\) can take any values for which \(f\) is a probability density function. Find the set of possible values of \(\mathrm{Var}(X)\).
64AAHL4-q00064.2 Presentation of dataPaper 2Easy8 marks
The times, \(t\) minutes, that 40 people spent in a waiting room are summarized in the table, where \(p\) and \(q\) are positive integers.
Time, \(t\) (min)
\(0\le t\lt 10\)
\(10\le t\lt 20\)
\(20\le t\lt 30\)
\(30\le t\lt 40\)
\(40\le t\lt 50\)
Frequency
5
\(p\)
12
\(q\)
4
Using mid-interval values, the estimated mean time is 25.25 minutes.
(a)
Show that \(p+q=19\).
(1)
(b)
Find the value of \(p\) and the value of \(q\).
(3)
(c)
Write down the modal class and the class that contains the median.
(2)
(d)
Find an estimate of the standard deviation of the times.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(5+p+12+q+4=40\)
A1
\(p+q=19\)
AG
(b)
\(\dfrac{5(5)+15p+25(12)+35q+45(4)}{40}=25.25\)
M1
\(15p+35q=505\)
A1
\(p=8,\ q=11\)
A1
(c)
modal class \(20\le t\lt 30\)
A1
median class \(20\le t\lt 30\)
A1
Cumulative frequencies 5, 13, 25: the 20th and 21st values lie in this class.
(d)
GDC with mid-interval values 5, 15, 25, 35, 45 and frequencies 5, 8, 12, 11, 4
(M1)
FT from their \(p,\ q\).
\(11.7\) min
A1
\(11.7233\ldots\)
Answers: (b) \(p=8,\ q=11\) · (c) both \(20\le t\lt 30\) · (d) 11.7 min
65AAHL4-q00104.3 Measures of central tendency and dispersionPaper 2Easy8 marks
The lengths, \(t\) minutes, of 100 phone calls made from an office are summarized below.
Length, \(t\) (min)
\(0\le t\lt 2\)
\(2\le t\lt 4\)
\(4\le t\lt 6\)
\(6\le t\lt 8\)
\(8\le t\lt 10\)
Number of calls
18
34
27
15
6
(a)
Write down the modal class.
(1)
(b)
Find an estimate of the mean and an estimate of the standard deviation of the call lengths.
(3)
(c)
The office’s phone records give the exact lengths of the 100 calls, which satisfy \(\sum t=421.6\) and \(\sum t^{2}=2310\). Using \(\sigma^{2}=\dfrac{\sum t^{2}}{n}-\bar{t}^{\,2}\), find the exact mean and the exact standard deviation of the call lengths.
(3)
(d)
Explain why the values found in part (b) differ from those found in part (c).
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(2\le t\lt 4\)
A1
(b)
mid-interval values 1, 3, 5, 7, 9 with the frequencies
the grouped estimates assume every call in a class lasts the mid-interval value; the actual lengths are spread within each class
R1
Answers: (a) \(2\le t\lt 4\) · (b) 4.14 min, 2.25 min · (c) 4.216 min, 2.31 min
66AAHL4-q00184.5 Sample spaces and probabilityPaper 2Easy8 marks
A computer chooses points \((x,\ y)\) uniformly at random in the unit square \(0\le x\le1,\ 0\le y\le1\).
(a)
Show that the probability that a point satisfies \(x^{2}+y^{2}\lt 1\) is \(\dfrac{\pi}{4}\).
(2)
(b)
In a simulation of 5000 points, 3921 satisfy \(x^{2}+y^{2}\lt 1\). Use this result to find an estimate of \(\pi\).
(2)
(c)
By finding the area of the region \(y\gt x^{2}\) inside the square, find the expected number of the 5000 points that satisfy \(y\gt x^{2}\).
(3)
(d)
Explain how the estimate in part (b) could be made more reliable.
(1)
Show mark scheme
Working
Mark
Notes
(a)
region is a quarter of a circle of radius 1, area \(\dfrac{\pi}{4}\); square has area 1
M1
probability \(=\dfrac{\pi/4}{1}\)
A1
\(=\dfrac{\pi}{4}\)
AG
(b)
\(\dfrac{\pi}{4}\approx\dfrac{3921}{5000}\)
(M1)
\(\pi\approx3.14\)
A1
\(3.1368\)
(c)
\(1-\displaystyle\int_{0}^{1}x^{2}\,\mathrm{d}x\)
(M1)
\(\dfrac{2}{3}\)
A1
\(5000\times\dfrac{2}{3}=3333.3\), about 3333
A1
(d)
use many more points: relative frequency gets closer to the theoretical probability as the number of trials increases
R1
Answers: (b) 3.14 · (c) about 3333
67AAHL4-q00244.6 Combined, conditional and independent eventsPaper 2Easy16 marks
The 400 students in Years 11 and 12 of a school were asked how they usually travel to school. The results are shown in the table.
Walk
Bus
Car
Total
Year 11
70
88
62
220
Year 12
50
72
58
180
Total
120
160
120
400
(a)
A stratified sample of 40 of these students, stratified by year group, is to be chosen. Find the number of Year 11 students and the number of Year 12 students in the sample.
(2)
(b)
A student is chosen at random from the 400. Write down the probability that the student travels by bus.
(1)
(c)(i)
Determine whether the events “travels by bus” and “is in Year 12” are independent.
(2)
(c)(ii)
Determine whether the events “walks” and “is in Year 12” are independent.
(2)
(d)
A student chosen at random does not walk. Find the probability that the student is in Year 11.
(2)
(e)
Three students are chosen at random from the 400, without replacement. Find the probability that all three walk.
(2)
(f)
Two Year 12 students are chosen at random, without replacement. Find the probability that they travel by different methods.
(3)
(g)
In another survey, the first 40 students to arrive at school one morning are questioned. Name this sampling method and give a reason why it may give a biased estimate of the proportion of students who travel by bus.
eg students arriving first may mostly come on the same bus (or are dropped off by car), so the sample is not representative of how all students travel
R1
Accept any reason linking arrival time to the method of travel.
Answers: (a) 22 and 18 · (b) 0.4 · (c)(i) independent (ii) not independent · (d) \(\tfrac{15}{28}\) · (e) 0.0265 · (f) 0.663 · (g) convenience
68AAHL4-q00364.8 Binomial distributionPaper 2Easy7 marks
Mira claims that she can predict the suit (hearts, diamonds, clubs or spades) of a playing card before it is turned over. To test her, a card is drawn at random from a full pack of 52 cards (13 of each suit) and she predicts its suit; the card is then replaced and the pack shuffled. This is done 48 times. Assume that Mira is only guessing, so that each prediction is correct with probability 0.25, independently. Let \(X\) be the number of correct predictions.
(a)
Find the mean and the standard deviation of \(X\).
(2)
(b)
Find \(\mathrm{P}(X\ge16)\).
(2)
(c)
The tester will accept Mira’s claim if she makes at least \(k\) correct predictions. Find the least value of \(k\) for which the probability that a person who is only guessing has her claim accepted is less than 0.01.
(2)
(d)
Explain why the card is replaced and the pack shuffled after each prediction.
so that the probability of a correct guess stays at 0.25 for every prediction and the predictions are independent (binomial conditions)
R1
Answers: (a) 12, 3 · (b) 0.123 · (c) \(k=20\)
69AAHL4-q00464.10 Regression line of x on yPaper 2Easy7 marks
A biologist records the air temperature, \(x\) °C, and the number of chirps, \(y\), made by a cricket in 15 seconds on eight evenings.
\(x\) (°C)
15
17
19
20
22
24
26
28
\(y\) (chirps)
22
27
29
33
35
40
42
47
The biologist wants to use these data to estimate the temperature from a count of chirps.
(a)
Find the equation of the regression line of \(x\) on \(y\).
(2)
(b)
Use your equation to estimate the temperature on an evening when a cricket makes 36 chirps in 15 seconds.
(2)
(c)
Interpret the gradient of your line in the context of the question.
(1)
(d)
On a cold evening a cricket makes 5 chirps in 15 seconds. Explain why the line should not be used to estimate the temperature on this evening.
(1)
(e)
Explain why the line found in part (a) should not be used to predict the number of chirps when the temperature is 21 °C.
(1)
Show mark scheme
Working
Mark
Notes
(a)
gradient \(0.533\)
A1
\(0.532667\ldots\)
intercept \(3.06\), giving \(x=0.533y+3.06\)
A1
\(3.06456\ldots\) Award A0A0 for \(y=1.86x-5.34\) (line of \(y\) on \(x\)).
(b)
\(x=0.532667(36)+3.06457\)
(M1)
\(22.2\) °C
A1
\(22.2405\ldots\) Accept 22.2–22.3 from rounded coefficients.
(c)
each extra chirp in 15 seconds corresponds to an increase in temperature of about 0.533 °C
A1
Must be in context and refer to the change in x per unit change in y.
(d)
5 is outside the range of the \(y\)-data (22 to 47), so it would be extrapolation
R1
(e)
to predict \(y\) from \(x\) the regression line of \(y\) on \(x\) must be used
R1
Answers: (a) \(x=0.533y+3.06\) · (b) 22.2 °C
70AAHL4-q00514.11 Conditional probability and independencePaper 2Easy7 marks
An archer shoots 8 arrows at a target. For each arrow, the probability that it hits the centre is 0.3, independently of all other arrows. Let \(X\) be the number of arrows that hit the centre.
(a)
Find \(\mathrm{P}(X=2)\).
(2)
(b)
Given that at least one arrow hits the centre, find the probability that exactly two arrows hit the centre.
(3)
(c)
Given that at least one arrow hits the centre, find the probability that the first arrow hits the centre.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(X\sim\mathrm{B}(8,\,0.3)\)
(M1)
\(0.296\)
A1
\(0.296475\ldots\)
(b)
\(\dfrac{\mathrm{P}(X=2)}{\mathrm{P}(X\ge1)}\)
(M1)
\(\mathrm{P}(X\ge1)=1-0.7^{8}=0.942352\)
(A1)
\(0.315\)
A1
\(0.314612\ldots\)
(c)
\(\dfrac{0.3}{0.942352}\)
(M1)
Recognizing that “first arrow hits” is a subset of “at least one hit”.
\(0.318\)
A1
\(0.318352\ldots\)
Answers: (a) 0.296 · (b) 0.315 · (c) 0.318
71AAHL4-q00594.12 Standardization and unknown normal parametersPaper 2Easy12 marks
The time, \(T\) minutes, for a ferry crossing is normally distributed with mean 48 minutes. 20% of crossings take more than 52 minutes.
(a)
Find the standard deviation of \(T\).
(3)
(b)
Find the probability that a crossing takes more than 55 minutes.
(1)
(c)
The ferry makes 6 crossings each day, with times independent of each other. Find the probability that on a particular day at least 2 crossings take more than 52 minutes.
(2)
(d)
Find the expected number of crossings that take more than 52 minutes in a period of 5 days.
(1)
(e)
Find the least number of crossings that must be made for the probability that at least one of them takes more than 55 minutes to exceed 0.95.
(3)
(f)
The company publishes a scheduled crossing time of \(s\) minutes, chosen so that only 5% of crossings take longer than \(s\) minutes. Find \(s\).
72AAHL4-q00724.13 Bayes’ theoremPaper 2Easy5 marks
In a climbing club, 60% of the members are experienced and 40% are novices. On each attempt at a difficult route, an experienced climber succeeds with probability 0.8 and a novice succeeds with probability 0.3, independently of other attempts. A member is chosen at random and attempts the route repeatedly until the first success.
(a)
Find the probability that an experienced climber succeeds for the first time on the third attempt.
(1)
(b)
Find the probability that the chosen member succeeds for the first time on the third attempt.
(2)
(c)
Given that the chosen member succeeds for the first time on the third attempt, find the probability that the member is experienced.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(0.2^{2}\times0.8=0.032\)
A1
(b)
\(0.6(0.032)+0.4(0.7^{2}\times0.3)\)
(M1)
\(0.078\)
A1
(c)
\(\dfrac{0.0192}{0.078}\)
(M1)
\(0.246\)
A1
\(\tfrac{16}{65}=0.246153\ldots\)
Answers: (a) 0.032 · (b) 0.078 · (c) 0.246
73AAHL4-q00984.14 Variance, continuous random variables and linear transformationsPaper 2Easy4 marks
The maximum daily temperature in a city in April, \(T\) °C, has mean \(15\) and standard deviation \(4\). The temperature in degrees Fahrenheit is \(F=1.8T+32\).
(a)
Find \(\mathrm{E}(F)\).
(2)
(b)
Find the standard deviation of \(F\) and \(\mathrm{Var}(F)\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(1.8(15)+32\)
(M1)
\(59\) °F
A1
(b)
\(\sigma_{F}=1.8\times4=7.2\)
A1
Award A0 for \(1.8\times4+32\).
\(\mathrm{Var}(F)=1.8^{2}\times16=51.84\)
A1
Answers: (a) 59 °F · (b) 7.2 °F, 51.84
74AAHL4-q01004.14 Variance, continuous random variables and linear transformationsPaper 2Easy6 marks
A bakery bakes 35 loaves of a speciality bread each day. The number of these loaves sold on a randomly chosen day, \(X\), has the distribution shown.
\(x\)
\(20\)
\(25\)
\(30\)
\(35\)
\(\mathrm{P}(X=x)\)
\(0.1\)
\(0.3\)
\(0.4\)
\(0.2\)
The daily profit, in euros, from this bread is \(D=3.2X-60\).
(a)
Find \(\mathrm{E}(X)\) and the standard deviation of \(X\).
(3)
(b)
Hence find the expected daily profit and the standard deviation of the daily profit.
(2)
(c)
Find the probability that the profit on a randomly chosen day exceeds €30.
75AAHL4-q01134.14 Variance, continuous random variables and linear transformationsPaper 2Easy15 marks
Three fair six-sided dice are rolled. Let \(D\) be the number of different scores shown. For example, the scores 4, 1, 4 give \(D=2\).
(a)
Show that \(\mathrm{P}(D=3)=\dfrac{5}{9}\).
(2)
(b)
Find \(\mathrm{P}(D=1)\) and \(\mathrm{P}(D=2)\).
(2)
(c)
Find \(\mathrm{E}(D)\) and \(\mathrm{Var}(D)\).
(3)
(d)
In a game, a player pays €2 to roll the three dice and receives \(12-4D\) euros. Find the mean and the standard deviation of the amount received, and hence state the expected gain of the player per game.
(3)
(e)
The game is played 30 times. Find the probability that all three scores are different in at least 20 of these games.
(3)
(f)
Given that at least one of the dice shows a 6, find the probability that \(D=1\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{6\times5\times4}{6^{3}}\) or \(1\times\dfrac{5}{6}\times\dfrac{4}{6}\)
76AAHL4-q00034.1 Sampling, bias and outliersPaper 2Medium10 marks
The times, in minutes, of 14 bus journeys on the same route were recorded.
Journey
1
2
3
4
5
6
7
8
9
10
11
12
13
14
Time (min)
12
14
15
15
16
17
17
18
19
20
21
23
29
52
Use the quartiles given by your GDC. An outlier is a value more than \(1.5\times\) IQR from the nearest quartile.
(a)
Find the lower quartile and the upper quartile of the times.
(2)
(b)
Show that 52 is an outlier but 29 is not.
(2)
(c)
The time of 52 minutes is removed from the data. Show that 29 is now an outlier.
(2)
(d)
Find the mean of the times before and after 52 is removed, and state, with a reason, whether the mean or the median is more affected by its removal.
(3)
(e)
The 52-minute journey took place on a day when a road was closed. Comment on whether the analyst should have removed it, and on whether 29 should then also be removed.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(Q_{1}=15\)
A1
\(Q_{3}=21\)
A1
(b)
\(\text{IQR}=6\); upper boundary \(21+1.5(6)=30\)
M1
\(52\gt 30\) so 52 is an outlier; \(29\lt 30\) so 29 is not
A1
(c)
new quartiles \(Q_{1}=15,\ Q_{3}=20.5\), IQR \(=5.5\)
M1
upper boundary \(20.5+1.5(5.5)=28.75\lt 29\), so 29 is now an outlier
A1
(d)
mean before \(=\dfrac{288}{14}=20.6\) min
A1
\(20.571\ldots\)
mean after \(=\dfrac{236}{13}=18.2\) min
A1
\(18.153\ldots\)
the mean is more affected (it falls by about 2.4 min, whereas the median only changes from 17.5 to 17)
R1
(e)
52 is a genuine value (a real journey) so it may be kept, or analysed separately with the reason stated; repeatedly removing the new largest value is not valid, so 29 should not be removed
R1
Accept any reasoned comment that distinguishes a genuine outlier from an error and rejects automatic repeated removal.
Answers: (a) \(Q_{1}=15,\ Q_{3}=21\) · (d) 20.6 min and 18.2 min
77AAHL4-q00074.2 Presentation of dataPaper 2Medium15 marks
The finishing times, \(t\) minutes, of 120 runners in a 5 km run are summarized below.
Time, \(t\) (min)
\(15\le t\lt 20\)
\(20\le t\lt 25\)
\(25\le t\lt 30\)
\(30\le t\lt 35\)
\(35\le t\lt 40\)
\(40\le t\lt 45\)
Number of runners
8
22
35
28
17
10
(a)
Write down the modal class.
(1)
(b)
Find an estimate of the mean and an estimate of the standard deviation of the times.
(3)
(c)
Construct a cumulative frequency table and use it, assuming the times are evenly spread within each class, to estimate the median time.
(3)
(d)
Estimate the number of runners who finished in less than 33 minutes.
(2)
(e)
From the raw data, the fastest time was 15.8 minutes, the slowest time 44.9 minutes, the lower quartile 24.2 minutes, the median 29.4 minutes and the upper quartile 34.6 minutes. Show that there are no outliers.
(3)
(f)
Use the five-number summary in part (e) to comment on whether the times could be normally distributed.
(1)
(g)
It is later found that every recorded time was 0.5 minutes too long. Write down the corrected estimates of the mean and the standard deviation.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(25\le t\lt 30\)
A1
(b)
mid-interval values 17.5, 22.5, …, 42.5 with the frequencies
60th value lies in \(25\le t\lt 30\): \(25+\dfrac{60-30}{35}\times5\)
(M1)
\(29.3\) min
A1
\(29.2857\ldots\) Accept from 29.2 to 29.4 (e.g. 60.5th value).
(d)
\(65+\dfrac{3}{5}\times28\)
(M1)
\(82\) runners
A1
\(81.8\)
(e)
IQR \(=34.6-24.2=10.4\); \(1.5\times10.4=15.6\)
M1
boundaries \(24.2-15.6=8.6\) and \(34.6+15.6=50.2\)
A1
\(15.8\gt 8.6\) and \(44.9\lt 50.2\), so there are no outliers
R1
(f)
the box is symmetric about the median (5.2 each side) and the whiskers are similar in length (8.4 and 10.3), so a normal distribution is plausible
R1
(g)
mean \(=29.25\) min
A1
FT from (b).
standard deviation unchanged, \(6.70\) min
A1
Answers: (a) \(25\le t\lt 30\) · (b) 29.75 min, 6.70 min · (c) 29.3 min · (d) 82 · (g) 29.25 min, 6.70 min
78AAHL4-q00134.4 Correlation and regression of y on xPaper 2Medium10 marks
The table shows the number of public electric-vehicle charging points, \(y\) thousand, in a region in the year \(t\). An analyst codes the years using \(u=\dfrac{t-2008}{2}\).
Year, \(t\)
2008
2010
2012
2014
2016
2018
2020
2022
\(y\) (thousands)
3.1
4.0
5.2
5.9
7.3
8.1
9.0
10.4
(a)
Find the equation of the regression line of \(y\) on \(u\).
(2)
(b)
Write down the value of Pearson’s product-moment correlation coefficient, \(r\), between \(y\) and \(t\).
(1)
(c)
Hence find the equation of the regression line of \(y\) on \(t\), in the form \(y=at+b\).
(3)
(d)
Interpret the value of \(a\) in context, and explain why \(b\) has no practical meaning.
(2)
(e)
Use your regression line to predict the number of charging points in 2030, and comment on the reliability of this prediction.
(2)
Show mark scheme
Working
Mark
Notes
(a)
gradient \(1.03\) (\(1.02619\ldots\))
A1
\(y=1.03u+3.03\)
A1
Intercept \(3.03333\ldots\)
(b)
\(r=0.998\)
A1
\(0.998126\ldots\); coding \(t\) linearly with a positive factor does not change \(r\).
\(-1027.26\ldots\) Accept \(-1027\) or \(-1028\); \(b\) is very sensitive to rounding of \(a\), so full accuracy must be carried. Also accept direct GDC regression on \(t\).
(d)
the number of charging points increases by about 513 per year
R1
\(b\) is the predicted value in year 0, far outside the data (and negative), so it is meaningless
R1
(e)
\(u=11\): \(y=14.3\), about 14 300 charging points
A1
\(14.32\ldots\)
unreliable: 2030 is outside the range of the data (extrapolation); growth may not stay linear
79AAHL4-q00154.4 Correlation and regression of y on xPaper 2Medium15 marks
A grower investigates the effect of a fertilizer on tomato plants. Eight plots are treated with different amounts of fertilizer, \(x\) grams per square metre, and the mean yield, \(y\) kg per plant, is recorded.
\(x\) (g m−2)
10
20
30
40
50
60
70
80
\(y\) (kg)
2.6
3.1
3.0
3.7
3.9
4.5
4.4
5.0
(a)
Find Pearson’s product-moment correlation coefficient, \(r\), and describe the correlation between \(x\) and \(y\).
(2)
(b)
Find the equation of the regression line of \(y\) on \(x\), in the form \(y=ax+b\).
(2)
(c)
Interpret the value of \(a\) in context.
(1)
(d)
Use your regression line to estimate the mean yield per plant when 35 g m−2 of fertilizer is used.
(2)
(e)(i)
Each square metre contains 3 plants. Tomatoes are sold for €1.20 per kg and fertilizer costs €0.08 per gram. Let \(V(x)\) be the value of the predicted yield of one square metre minus the cost of the fertilizer used on it, in euros.
Show that \(V(x)=0.04x+8.19\).
(2)
(e)(ii)
The grower decides to use the amount of fertilizer, within the range of the data, that maximizes \(V(x)\). State this amount, and explain why the grower should not conclude that using even more fertilizer would be better.
(2)
(f)
When the amount found in part (e)(ii) is used, the yield of a plant is normally distributed with mean equal to the value predicted by the regression line and standard deviation 0.6 kg. Find the probability that a randomly chosen plant yields less than 4 kg.
(2)
(g)
Ten such plants are chosen at random. Find the probability that at most one of them yields less than 4 kg.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.978\)
A1
\(0.978399\ldots\)
strong positive (linear) correlation
A1
(b)
\(a=0.0333\)
A1
\(0.03333\ldots\)
\(b=2.28\); \(y=0.0333x+2.28\)
A1
\(2.275\) Award A1A0 if not written as an equation.
(c)
for each extra gram per square metre of fertilizer, the mean yield increases by about 0.0333 kg per plant
A1
Must refer to an increase of 1 unit in x.
(d)
\(0.033333(35)+2.275\)
(M1)
\(3.44\) kg
A1
\(3.44166\ldots\)
(e)(i)
\(V(x)=3\times1.20\times(0.033333x+2.275)-0.08x\)
M1
\(=0.12x+8.19-0.08x\)
A1
\(3.6\times\tfrac{1}{30}=0.12\) and \(3.6\times2.275=8.19\).
\(V(x)=0.04x+8.19\)
AG
(e)(ii)
\(80\) g m−2 (the largest amount in the data, since \(V\) is increasing)
A1
the linear model is only supported for 10 ≤ x ≤ 80; beyond this it is extrapolation and the yield may level off (or fall), so V may not keep increasing
R1
(f)
\(Y\sim\mathrm{N}(4.94167,\,0.6^{2})\), where \(\tfrac{80}{30}+2.275=4.94166\ldots\); \(\mathrm{P}(Y\lt4)\)
(M1)
\(0.0583\)
A1
\(0.0582722\ldots\) Accept 0.0582–0.0586 from a mean rounded to 4.94.
80AAHL4-q00234.6 Combined, conditional and independent eventsPaper 2Medium7 marks
A contestant answers three questions in a quiz. The probability that she answers the first question correctly is 0.7. For each later question, the probability of a correct answer is 0.8 if the previous answer was correct and 0.5 if the previous answer was wrong.
(a)
Find the probability that she answers all three questions correctly.
(2)
(b)
Find the probability that she answers exactly two of the questions correctly.
(3)
(c)
Given that she answers exactly two questions correctly, find the probability that she answered the first question correctly.
81AAHL4-q00344.8 Binomial distributionPaper 2Medium8 marks
Each day a machine makes 20 components. Each component is defective with probability 0.06, independently of all others. A day is classed as “bad” if 3 or more of that day’s components are defective.
(a)
Find the probability that a given day is bad.
(2)
(b)
Find the probability that, in a five-day week, there are at least two bad days.
(3)
(c)
Find the expected number of bad days in a year of 250 working days.
(2)
(d)
State one assumption needed for your answer to part (b).
whether a day is bad is independent of other days (and the defect rate stays at 0.06 every day)
R1
Answers: (a) 0.115 · (b) 0.104 · (c) 28.7
82AAHL4-q00354.8 Binomial distributionPaper 2Medium7 marks
An archer shoots 10 arrows. Each arrow hits the target with probability \(p\), independently. Let \(X\) be the number of arrows that hit the target. The variance of \(X\) is 2.1.
(a)
Show that \(p^{2}-p+0.21=0\).
(2)
(b)
Find the two possible values of \(p\).
(1)
(c)
It is known that \(\mathrm{P}(X\ge6)\gt 0.5\). Determine the value of \(p\).
(2)
(d)
For this value of \(p\), find the probability that the number of hits equals its expected value.
83AAHL4-q00384.8 Binomial distributionPaper 2Medium16 marks
A laboratory tests blood samples for a rare condition. Each person has the condition with probability 0.04, independently of all other people. To save tests, the laboratory uses pooled testing.
The samples of a group of 8 people are mixed and the mixture is tested once. The mixture tests positive if and only if at least one person in the group has the condition.
If the mixture tests negative, no further tests are needed for the group.
If the mixture tests positive, each of the 8 samples is then tested individually.
Let \(N\) be the total number of tests carried out for a group of 8 people.
(a)
Find the probability that the mixture tests positive.
(2)
(b)
Show that \(\mathrm{E}(N)=3.23\), correct to three significant figures.
(3)
(c)
Find \(\mathrm{Var}(N)\).
(3)
(d)
Each test costs €12 and the laboratory also pays €30 to process each group. Find the mean and the standard deviation of the total cost, in euros, of testing a group.
(3)
(e)
Given that the mixture for a group tests positive, find the probability that exactly one person in the group has the condition.
(2)
(f)(i)
Find the expected number of tests per person under this scheme, and hence the percentage reduction in the expected number of tests compared with testing every person individually.
(2)
(f)(ii)
Explain why pooled testing would save fewer tests if the condition were much more common.
(1)
Show mark scheme
Working
Mark
Notes
(a)
\(1-\mathrm{P}(\text{nobody has the condition})=1-0.96^{8}\)
(M1)
Or \(X\sim\mathrm{B}(8,\,0.04)\), \(\mathrm{P}(X\ge1)\).
\(0.279\)
A1
\(0.278610\ldots\)
(b)
\(N=1\) with probability \(0.721389\ldots\) and \(N=9\) with probability \(0.278610\ldots\)
84AAHL4-q00394.8 Binomial distributionPaper 2Medium16 marks
A test has 40 multiple-choice questions, each with four options of which exactly one is correct. For each question, independently, Sami knows the answer with probability 0.6; otherwise he chooses one of the four options at random.
(a)
Show that the probability that Sami answers a question correctly is 0.7.
(2)
(b)
Given that Sami answers two particular questions correctly, find the probability that he knew the answers to both of them.
(2)
(c)
Let \(X\) be the number of questions Sami answers correctly. Find \(\mathrm{E}(X)\) and \(\mathrm{Var}(X)\).
(2)
(d)
Each correct answer scores 4 marks and each wrong answer scores \(-1\) mark. Let \(S\) be Sami’s total score. Show that \(S=5X-40\), and hence find the mean and the standard deviation of \(S\).
(3)
(e)
Find \(\mathrm{P}(S\ge120)\).
(3)
(f)
Rhea knows none of the answers and guesses every question at random. Find the probability that her total score is positive.
(2)
(g)
The test designer wants the expected score for a random guess on a single question to be zero. Keeping 4 marks for a correct answer, find the number of marks that should be deducted for a wrong answer.
85AAHL4-q00414.9 The normal distributionPaper 2Medium7 marks
At an airport, the time, \(W\) minutes, that a passenger waits for the shuttle bus is modelled by a normal distribution with mean 6.5 minutes and standard deviation 4 minutes.
(a)
Use this model to find the probability that a passenger waits more than 10 minutes.
(2)
(b)(i)
Use this model to find \(\mathrm{P}(W\lt 0)\).
(1)
(b)(ii)
Hence comment on the suitability of this model.
(1)
(c)
The bus company replaces the model by \(W\sim\mathrm{N}(6.5,\,\sigma^{2})\), choosing \(\sigma\) so that \(\mathrm{P}(W\lt 0)=0.001\). Find the value of \(\sigma\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(W\gt 10)\) with \(W\sim\mathrm{N}(6.5,\,4^{2})\), e.g. sketch or GDC input
(M1)
\(0.191\)
A1
\(0.190786\ldots\)
(b)(i)
\(0.0521\)
A1
\(0.0520812\ldots\)
(b)(ii)
a waiting time cannot be negative, but the model gives about 5% of passengers a negative waiting time, so the model is not suitable (for short waiting times)
R1
Must refer to the size of the probability from (b)(i) and to waiting times not being negative.
(c)
\(\dfrac{0-6.5}{\sigma}=z\) where \(\mathrm{P}(Z\lt z)=0.001\)
86AAHL4-q00434.9 The normal distributionPaper 2Medium8 marks
The volume, \(V\) ml, of juice in a carton is normally distributed with mean 502 ml and standard deviation 1.8 ml. Each carton is labelled “500 ml”. Cartons are sold in packs of five, and the volumes in different cartons are independent.
(a)
Find the probability that a randomly chosen carton contains less than 500 ml.
(2)
(b)
Find the probability that none of the cartons in a pack contains less than 500 ml.
(2)
(c)
The manufacturer adjusts the mean volume to \(\mu\) ml, keeping the standard deviation at 1.8 ml, so that the probability that none of the cartons in a pack contains less than 500 ml is 0.9. Find the value of \(\mu\). Give your answer correct to two decimal places.
(4)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(V\lt 500)\) with \(V\sim\mathrm{N}(502,\,1.8^{2})\)
(M1)
\(0.133\)
A1
\(0.133260\ldots\)
(b)
\((1-0.133260\ldots)^{5}\)
(M1)
Or binomial \(\mathrm{P}(Y=0)\) with \(Y\sim\mathrm{B}(5,\,0.13326)\).
\(0.489\)
A1
\(0.489151\ldots\)
(c)
\(\big(\mathrm{P}(V\ge 500)\big)^{5}=0.9\)
M1
\(\mathrm{P}(V\ge 500)=0.9^{1/5}=0.979148\ldots\)
A1
\(\dfrac{500-\mu}{1.8}=-2.03647\ldots\)
M1
Standardizing with the \(z\)-value for a lower tail of \(0.020851\ldots\)
\(\mu=503.67\) ml
A1
\(503.6656\ldots\) Award A0 for 503.66 (truncation) or 503.7 (not 2 d.p.).
Answers: (a) 0.133 · (b) 0.489 · (c) \(\mu=503.67\) ml
87AAHL4-q00454.9 The normal distributionPaper 2Medium16 marks
The error, \(X\) mm, in a length measured by a laser device is modelled by the probability density function
Write down \(\mathrm{E}(X)\), giving a reason for your answer.
(1)
(c)
Show that \(\mathrm{Var}(X)=\dfrac{1}{5}\).
(3)
(d)
Find \(\mathrm{P}(|X|\lt0.5)\).
(2)
(e)
An engineer instead models the error by \(Y\sim\mathrm{N}\left(0,\,\tfrac{1}{5}\right)\), which has the same mean and variance as \(X\). Find \(\mathrm{P}(|Y|\lt0.5)\).
(2)
(f)
Find the value of \(t\) such that \(\mathrm{P}(X\gt t)=0.05\), and the value of \(s\) such that \(\mathrm{P}(Y\gt s)=0.05\).
(4)
(g)
Find \(\mathrm{P}(|Y|\gt1)\), and hence comment on the suitability of the normal model.
Or \(\dfrac{3}{4}\left(\dfrac{2}{3}-t+\dfrac{t^{3}}{3}\right)=0.05\).
solving with GDC
(M1)
\(t=0.729\)
A1
\(0.729299\ldots\)
\(s=0.736\)
A1
\(0.735600\ldots\) (inverse normal).
(g)
\(\mathrm{P}(|Y|\gt1)=0.0253\)
A1
\(0.0253473\ldots\)
under the original model errors larger than 1 mm are impossible, but the normal model gives them a probability of about 2.5%, so it overestimates the tails (although it is close for central probabilities)
R1
Accept any comment comparing the bounded and unbounded models.
88AAHL4-q00474.10 Regression line of x on yPaper 2Medium15 marks
In a road-safety study, nine drivers were tested at different times after waking. For each driver the number of hours, \(x\), since waking and the reaction time, \(y\) milliseconds (ms), were recorded.
\(x\) (hours)
2
4
6
8
10
12
14
16
18
\(y\) (ms)
246
251
260
258
274
279
291
298
311
(a)
Find the value of Pearson’s product-moment correlation coefficient, \(r\), and describe the correlation.
(2)
(b)
Find the equation of the regression line of \(y\) on \(x\), and use it to estimate the reaction time of a driver who has been awake for 11 hours.
(3)
(c)
A driver involved in an accident has a reaction time of 295 ms. Find the equation of the regression line of \(x\) on \(y\), and use it to estimate the number of hours since this driver woke.
(3)
(d)
Explain why the line of \(x\) on \(y\), rather than a rearrangement of the line in part (b), is used in part (c).
(1)
(e)
The reaction times of well-rested drivers are normally distributed with mean 255 ms and standard deviation 28 ms. Find the probability that a well-rested driver has a reaction time greater than 320 ms.
(2)
(f)
Find the reaction time, \(t\) ms, such that only 2% of well-rested drivers have a reaction time greater than \(t\).
(2)
(g)
Fifteen well-rested drivers are chosen at random. Find the probability that at least one of them has a reaction time greater than \(t\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(r=0.985\)
A1
\(0.985064\ldots\)
strong positive (linear) correlation
A1
FT their \(r\).
(b)
\(y=4.03x+234\)
A1
\(4.03333\ldots,\ 233.888\ldots\)
\(y=4.03333(11)+233.889\)
(M1)
\(278\) ms
A1
\(278.255\ldots\)
(c)
\(x=0.241y-56.0\)
A1
\(0.240583\ldots,\ -55.9732\ldots\)
\(x=0.240583(295)-55.9733\)
(M1)
\(15.0\) hours
A1
\(14.9987\ldots\) Accept 14.9–15.1 from rounded coefficients. Award (M1)A0 for 15.2 from rearranging the \(y\) on \(x\) line.
(d)
\(x\) is being estimated from a known value of \(y\), and the line of \(y\) on \(x\) minimizes errors in \(y\), not in \(x\)
R1
(e)
\(\mathrm{P}(Y\gt 320)\) with \(Y\sim\mathrm{N}(255,\,28^{2})\)
(M1)
\(0.0101\)
A1
\(0.0101318\ldots\)
(f)
\(\mathrm{P}(Y\gt t)=0.02\), e.g. invNorm(0.98, 255, 28)
(M1)
\(t=313\)
A1
\(312.504\ldots\) Accept 312.5.
(g)
\(1-0.98^{15}\)
(M1)
Or \(1-\mathrm{P}(N=0)\) with \(N\sim\mathrm{B}(15,\,0.02)\).
89AAHL4-q00554.12 Standardization and unknown normal parametersPaper 2Medium7 marks
The time, \(T\) minutes, that a worker takes to assemble a flat-pack desk is normally distributed with mean \(\mu\) and standard deviation \(\sigma\). It is known that 30% of workers take less than 20 minutes and 50% of workers take between 20 and 30 minutes.
(a)
Find the value of \(\mu\) and the value of \(\sigma\).
(4)
(b)
After a training course the mean time is unchanged, but the standard deviation changes so that 60% of workers take between 20 and 30 minutes. Find the new standard deviation.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(T\lt 30)=0.3+0.5=0.8\)
(M1)
\(z\)-values \(-0.524400\ldots\) and \(0.841621\ldots\)
(A1)
\(\dfrac{20-\mu}{\sigma}=-0.524400\) and \(\dfrac{30-\mu}{\sigma}=0.841621\)
M1
Award M0 if 0.5 is used as the cumulative probability for 30.
\(\mu=23.8,\ \sigma=7.32\)
A1
\(23.8388\ldots,\ 7.32052\ldots\) Both values required.
(b)
\(\mathrm{P}(20\lt T\lt 30)=0.6\) with \(T\sim\mathrm{N}(23.8388,\,s^{2})\)
M1
FT their \(\mu\).
use of GDC to solve, e.g. graph of \(y=\mathrm{normalcdf}(20,\,30,\,23.8388,\,s)\) intersecting \(y=0.6\)
(M1)
The interval is not symmetric about the mean, so a single \(z\)-value cannot be used.
\(s=5.82\)
A1
\(5.82458\ldots\) Accept 5.81–5.83 (5.816 from \(\mu=23.8\)).
90AAHL4-q00574.12 Standardization and unknown normal parametersPaper 2Medium8 marks
The diameters, \(D\) mm, of bolts made by a machine are normally distributed with mean 12 mm and standard deviation \(\sigma\) mm. A bolt is rejected if its diameter differs from 12 mm by more than 0.05 mm. Exactly 4% of bolts are rejected.
(a)
Find the value of \(\sigma\).
(3)
(b)
The mean diameter drifts to 12.02 mm, while \(\sigma\) is unchanged. Find the percentage of bolts that are now rejected.
(2)
(c)
Explain why the percentage rejected increases, although the width of the accepted interval and the standard deviation are unchanged.
(1)
(d)
With the mean at 12.02 mm, a bolt is rejected. Find the probability that it was rejected for being too large.
(2)
Show mark scheme
Working
Mark
Notes
(a)
by symmetry \(\mathrm{P}(D\lt 11.95)=0.02\)
M1
\(\dfrac{11.95-12}{\sigma}=-2.05374\ldots\)
(A1)
\(\sigma=0.0243\)
A1
\(0.0243457\ldots\)
(b)
\(\mathrm{P}(D\lt 11.95)+\mathrm{P}(D\gt 12.05)\) with \(D\sim\mathrm{N}(12.02,\,0.0243457^{2})\)
(M1)
\(11.1\%\)
A1
\(0.110946\ldots\) Accept 11.0%–11.2%.
(c)
the upper tail beyond 12.05 gains more probability than the lower tail below 11.95 loses, because the normal curve is highest near the mean
91AAHL4-q00714.13 Bayes’ theoremPaper 2Medium9 marks
A missing hiker is in exactly one of three regions, \(R_{1}\), \(R_{2}\) and \(R_{3}\), with probabilities 0.45, 0.35 and 0.20 respectively. If the hiker is in the region being searched, a search of that region finds the hiker with probability 0.7 for \(R_{1}\), 0.85 for \(R_{2}\) and 0.9 for \(R_{3}\). A search never finds the hiker in a region where the hiker is not. Region \(R_{1}\) is searched first.
(a)
Find the probability that the first search does not find the hiker.
(2)
(b)
Given that the first search does not find the hiker, find the probability that the hiker is in each of the three regions.
(3)
(c)
The rescue team next searches the region in which the hiker is now most likely to be. Write down this region.
(1)
(d)
The second search also does not find the hiker. Find the probability that the hiker is in region \(R_{3}\).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(0.45(0.3)+0.35+0.20\)
(M1)
Recognizing that the hiker is not found if in \(R_{1}\) and missed, or in \(R_{2}\) or \(R_{3}\).
92AAHL4-q00764.13 Bayes’ theoremPaper 2Medium15 marks
The emails received by a company are of three types: 25% are spam (\(S\)), 35% are promotional (\(R\)) and 40% are personal (\(P\)). An email filter flags an email as suspicious with the following probabilities.
Type of email
Spam
Promotional
Personal
P(flagged)
0.92
0.15
0.02
An email is chosen at random.
(a)
Find the probability that the email is flagged.
(2)
(b)
Given that the email is flagged, find the probability that it is spam.
(2)
(c)
Given that the email is not flagged, find the probability that it is personal.
(3)
(d)
On one day the company receives 200 emails, independently of each other. Find the expected number of personal emails that are flagged, and the probability that at least 3 personal emails are flagged.
(4)
(e)
Two emails, chosen independently at random, are both flagged. Find the probability that they are of the same type.
(3)
(f)
Suggest a reason why the assumption in part (d) that the emails are independent of each other may not be valid.
93AAHL4-q00964.14 Variance, continuous random variables and linear transformationsPaper 2Medium8 marks
A bag contains 3 red and 5 blue counters. In a game, a player pays a fee and then takes two counters at random from the bag, without replacement. The player receives €7 for each red counter taken. Let \(R\) be the number of red counters taken.
(a)
Show that \(\mathrm{P}(R=1)=\dfrac{15}{28}\).
(2)
(b)
Find the fee that would make the game fair.
(2)
(c)
Find the standard deviation of the amount the player receives.
(2)
(d)
The fee is set at €6. Find the expected total gain of a player who plays the game 50 times.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac38\times\dfrac57+\dfrac58\times\dfrac37\)
M1
Or \(\dfrac{\binom31\binom51}{\binom82}\). Must see both orders, or a combination count.
game A, because it has the same mean but a smaller variance (standard deviation) than game B
R1
Must refer to the variance or standard deviation. FT their values.
Answers: (a) \(k=13\) · (b) \(5.04\) · (c) \(4.2\) · (d) game A (smaller spread)
95AAHL4-q00994.14 Variance, continuous random variables and linear transformationsPaper 2Medium6 marks
The raw mark \(X\) of a randomly chosen candidate in a test has \(\mathrm{E}(X)=54\) and standard deviation \(12.5\). The marks are rescaled using \(Y=aX+b\), where \(a\gt 0\), so that the scaled marks have mean \(65\) and standard deviation \(10\).
(a)
Find the value of \(a\) and the value of \(b\).
(3)
(b)
Find the scaled mark of a candidate whose raw mark is \(72\).
(1)
(c)
A candidate has a scaled mark of \(50\). Find this candidate’s raw mark.
96AAHL4-q01014.14 Variance, continuous random variables and linear transformationsPaper 2Medium8 marks
The continuous random variable \(X\) has probability density function \(f(x)=k\sqrt{9-x^{2}}\) for \(0\le x\le3\), and \(f(x)=0\) otherwise, where \(k\) is a positive constant.
(a)
By considering the shape of the graph of \(y=\sqrt{9-x^{2}}\) for \(0\le x\le3\), show that \(k=\dfrac{4}{9\pi}\).
(2)
(b)
Find \(\mathrm{E}(X)\).
(2)
(c)
Find the standard deviation of \(X\).
(2)
(d)
Find the median of \(X\).
(2)
Show mark scheme
Working
Mark
Notes
(a)
the graph is a quarter of the circle \(x^{2}+y^{2}=9\), so the area under it is \(\dfrac14\pi(3)^{2}=\dfrac{9\pi}{4}\)
\(1.21191\ldots\) Award A0 for values outside \([0,3]\).
Answers: (b) 1.27 · (c) 0.793 · (d) 1.21
97AAHL4-q01024.14 Variance, continuous random variables and linear transformationsPaper 2Medium9 marks
The delay, \(X\) hours, of a parcel delivery is modelled by the probability density function \(f(x)=k\mathrm{e}^{-(x-1)^{2}}\) for \(0\le x\le4\), and \(f(x)=0\) otherwise, where \(k\) is a positive constant.
(a)
Find the value of \(k\).
(2)
(b)
Use differentiation to show that the mode of \(X\) is \(1\).
(2)
(c)
Find \(\mathrm{E}(X)\) and the standard deviation of \(X\).
Find the mean and the standard deviation of \(T\).
(4)
(e)
Find the probability that a customer waits more than 4 minutes.
(2)
(f)
Leila uses the taxi rank on 10 days. Her waiting times are independent. Find the probability that she waits more than 4 minutes on at least 3 of these days.
\(\displaystyle\int_{0}^{2}\sin\left(\dfrac{\pi x}{4}\right)\mathrm{d}x=\left[-\dfrac{4}{\pi}\cos\dfrac{\pi x}{4}\right]_{0}^{2}=\dfrac{4}{\pi}\) and \(\displaystyle\int_{2}^{5}\dfrac{5-x}{3}\,\mathrm{d}x=\dfrac{3}{2}\)
A1
Second integral may be found as the area of a triangle.
\(f\) is increasing on \(0\le x\le2\) (sine increasing for \(0\le\frac{\pi x}{4}\le\frac{\pi}{2}\)) and decreasing on \(2\lt x\le5\), so the maximum of \(f\) is at \(x=2\)
R1
(c)
\(\dfrac{4a}{\pi}\) or \(\displaystyle\int_{0}^{2}f(x)\,\mathrm{d}x\)
(M1)
\(a=0.360589\ldots\)
\(0.459\)
A1
\(0.459116\ldots\) Accept \(\dfrac{8}{8+3\pi}\).
(d)
\(\mathrm{P}(X\le2)\lt0.5\), so the median \(m\) lies in \(2\lt m\le5\)
100AAHL4-q00144.4 Correlation and regression of y on xPaper 2Hard10 marks
A school offers extra tutoring. Group P is a support group of students with low prior attainment; group Q are the other students. The table and scatter diagram show the hours of tutoring, \(x\), and the score in a test, \(y\), for 14 students.
Group P: \(x\)
6
7
8
9
10
11
12
Group P: \(y\)
38
44
41
49
52
50
57
Group Q: \(x\)
0
1
2
3
4
5
6
Group Q: \(y\)
63
70
66
74
72
79
81
(a)
Find the value of Pearson’s product-moment correlation coefficient, \(r\), for group P and for group Q separately.
(2)
(b)
Find the value of \(r\) for all 14 students together.
(2)
(c)
The head teacher looks only at the combined data and concludes that tutoring lowers test scores. Comment on this conclusion.
(2)
(d)
A student in group Q received 4.5 hours of tutoring. Find the equation of an appropriate regression line and use it to estimate this student’s score.
(3)
(e)
Explain why the regression line for all 14 students should not be used in part (d).
(1)
Show mark scheme
Working
Mark
Notes
(a)
group P: \(r=0.931\)
A1
\(0.93149\ldots\)
group Q: \(r=0.923\)
A1
\(0.92335\ldots\)
(b)
entering all 14 pairs
(M1)
\(r=-0.521\)
A1
\(-0.52111\ldots\)
(c)
within each group there is strong positive correlation, so the data do not support the claim
R1
the negative overall correlation is caused by group membership (prior attainment): weaker students received more tutoring; correlation does not imply causation
R1
(d)
regression of \(y\) on \(x\) using group Q only
(M1)
\(y=2.79x+63.8\)
A1
\(2.78571\ldots,\ 63.7857\ldots\)
\(76.3\)
A1
\(76.32\ldots\) Accept 76.
(e)
the combined data are two separate clusters with different relationships, so the combined line (gradient \(-2\)) does not model either group
101AAHL4-q00374.8 Binomial distributionPaper 2Hard15 marks
A shop receives large batches of light bulbs. In a batch, a proportion \(p\) of bulbs is defective, and bulbs are defective independently of each other. The shop considers the following plans for deciding whether to accept a batch.
Plan A: test 20 bulbs; accept the batch if at most one of them is defective.
Plan B: test 30 bulbs; accept the batch if at most two of them are defective.
(a)
For Plan A with \(p=0.05\), find the probability that the batch is accepted.
(2)
(b)
Show that, under Plan A, the probability that a batch is accepted is \((1-p)^{19}(1+19p)\).
(2)
(c)
Find the value of \(p\) for which Plan A accepts a batch with probability 0.5.
(2)
(d)
For Plan B, find the probability that the batch is accepted when \(p=0.05\) and when \(p=0.15\).
(3)
(e)
The shop requires a plan that accepts a good batch (\(p=0.05\)) with probability at least 0.8 and accepts a poor batch (\(p=0.15\)) with probability at most 0.2. Determine which of Plans A and B meet both requirements.
(2)
(f)
Plan C is to test \(n\) bulbs and accept the batch only if none of them is defective. Show that no value of \(n\) allows Plan C to meet both requirements in part (e).
Plan A fails: \(0.736\lt 0.8\) (for \(p=0.15\) it accepts with probability \(0.176\))
R1
\(0.175557\ldots\)
Plan B meets both: \(0.812\ge0.8\) and \(0.151\le0.2\)
A1
FT from (a) and (d).
(f)
\(0.95^{n}\ge0.8\Rightarrow n\le\dfrac{\ln0.8}{\ln0.95}=4.35\), so \(n\le4\)
M1
\(0.85^{n}\le0.2\Rightarrow n\ge\dfrac{\ln0.2}{\ln0.85}=9.90\), so \(n\ge10\)
M1
no integer satisfies both \(n\le4\) and \(n\ge10\), so Plan C cannot meet both requirements
R1
(g)
Plan A tests only 20 bulbs rather than 30, so it is cheaper/quicker
R1
Answers: (a) 0.736 · (c) \(p=0.0825\) · (d) 0.812 and 0.151 · (e) Plan B only · (f) \(n\le4\) and \(n\ge10\) incompatible
102AAHL4-q00424.9 The normal distributionPaper 2Hard6 marks
The random variable \(X\) is normally distributed with mean 40 and standard deviation 6. The constant \(k\gt 0\) is such that \(\mathrm{P}(40-k\lt X\lt 40+2k)=0.75\).
attempt to set up an equation in \(k\), e.g. \(\mathrm{P}(X\lt 40+2k)-\mathrm{P}(X\lt 40-k)=0.75\), or \(\Phi\!\left(\tfrac{2k}{6}\right)-\Phi\!\left(-\tfrac{k}{6}\right)=0.75\)
M1
Award M0 for treating the interval as symmetric about the mean.
use of GDC to solve, e.g. graph of \(y=\mathrm{normalcdf}(40-k,\,40+2k,\,40,\,6)\) intersecting \(y=0.75\)
Recognising that \(\{X\gt 40+2k\}\cap\{X\gt 40-k\}=\{X\gt 40+2k\}\).
\(\mathrm{P}(X\gt 50.0)=0.0477\) and \(\mathrm{P}(X\gt 35.0)=0.798\)
A1
Both values. FT their \(k\).
\(0.0598\)
A1
\(0.0598455\ldots\) Accept 0.0597–0.0599.
Answers: (a) \(k=5.00\) · (b) 0.0598
103AAHL4-q00444.9 The normal distributionPaper 2Hard14 marks
A machine fills bags of coffee. The mass, \(X\) grams, of coffee in a bag is normally distributed with mean \(\mu\) grams and standard deviation 4 grams. The value of \(\mu\) can be set by the operator.
The coffee costs the company €0.02 per gram. Every bag is checked: a bag with \(X\ge245\) is sold for €8, and a bag with \(X\lt245\) is rejected and cannot be sold. Let \(G\) be the profit, in euros, on a bag, so that \(G=8-0.02X\) if the bag is sold and \(G=-0.02X\) if it is rejected.
(a)
For \(\mu=250\), find the probability that a bag is rejected.
(2)
(b)
Show that \(\mathrm{E}(G)=8\,\mathrm{P}(X\ge245)-0.02\mu\).
(2)
(c)
Find \(\mathrm{E}(G)\) when \(\mu=250\).
(2)
(d)
Find the value of \(\mu\) that maximizes \(\mathrm{E}(G)\), and the maximum value of \(\mathrm{E}(G)\).
(3)
(e)
For the value of \(\mu\) found in part (d), find the probability that a bag is rejected, and hence the expected number of rejected bags in a run of 10 000 bags.
(2)
(f)
A regulation states that at most 1% of bags may contain less than 245 g. Find the least value of \(\mu\) that satisfies the regulation, and determine whether the value found in part (d) satisfies it.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(X\lt245)\) with \(X\sim\mathrm{N}(250,\,4^{2})\)
(M1)
\(0.106\)
A1
\(0.105649\ldots\)
(b)
\(G=-0.02X+B\), where \(B=8\) if the bag is sold and \(0\) otherwise, so \(\mathrm{E}(B)=8\,\mathrm{P}(X\ge245)\)
M1
Or an equivalent argument splitting the expectation.
104AAHL4-q00524.11 Conditional probability and independencePaper 2Hard10 marks
The masses, \(M\) kg, of watermelons from a farm are normally distributed with mean 6.2 kg and standard deviation 1.1 kg. A watermelon with a mass greater than 7 kg is sold as “premium”.
(a)
Find the probability that a randomly chosen watermelon is premium.
(2)
(b)
A premium watermelon is chosen at random. Find the probability that its mass is greater than 8 kg.
(3)
(c)
Find the mass \(m\) kg such that half of all premium watermelons have a mass greater than \(m\) kg.
(3)
(d)
Two premium watermelons are chosen at random. Find the probability that exactly one of them has a mass greater than 8 kg.
105AAHL4-q00564.12 Standardization and unknown normal parametersPaper 2Hard9 marks
A random variable \(X\) is normally distributed with mean \(\mu\) and variance \(\mu\), where \(\mu\gt 0\). It is known that \(\mathrm{P}(X\gt 10)=0.05\).
(a)
Let \(s=\sqrt{\mu}\). Show that \(s^{2}+as-10=0\), where \(a=1.64\) correct to three significant figures.
(2)
(b)
Hence find the value of \(\mu\).
(3)
(c)
Find \(\mathrm{P}(X\lt 4)\).
(2)
(d)
\(X\) models a quantity that cannot be negative. Find \(\mathrm{P}(X\lt 0)\) and comment on whether this makes the model unreasonable.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\dfrac{10-\mu}{\sqrt{\mu}}=1.64485\ldots\)
M1
Standard deviation \(\sqrt{\mu}\) and the \(z\)-value for an upper tail of 0.05.
\(10-s^{2}=1.64485s\)
A1
\(s^{2}+1.64s-10=0\)
AG
(b)
solving \(s^{2}+1.64485s-10=0\), e.g. quadratic formula or GDC
M1
\(s=2.44504\ldots\), rejecting \(s=-4.08990\ldots\) because \(s=\sqrt{\mu}\gt 0\)
R1
\(\mu=5.98\)
A1
\(5.97826\ldots\) Accept 5.97–5.99.
(c)
\(X\sim\mathrm{N}(5.97826,\ 5.97826)\), i.e. \(\sigma=2.44504\)
(M1)
Award M0 for using 5.97826 as the standard deviation.
\(0.209\)
A1
\(0.209232\ldots\)
(d)
\(\mathrm{P}(X\lt 0)=0.00724\)
A1
\(0.00724166\ldots\)
this probability is very small (less than 1%), so the model is not made unreasonable
R1
FT their value; the comment must refer to the size of the probability.
106AAHL4-q00584.12 Standardization and unknown normal parametersPaper 2Hard15 marks
On a beach, 70% of the crabs are of species \(A\) and 30% are of species \(B\). The carapace width, \(W\) mm, of a crab is normally distributed within each species.
For species \(A\), 10% of crabs have \(W\gt60\) and 25% have \(W\lt50\).
For species \(B\), the mean width is 64 mm and 15% of crabs have \(W\lt56\).
(a)
Find the mean and the standard deviation of the widths of species \(A\).
(5)
(b)
Find the standard deviation of the widths of species \(B\).
(2)
(c)
A crab is chosen at random from the beach. Find the probability that its width is greater than 60 mm.
(3)
(d)
A crab chosen at random has a width greater than 60 mm. Find the probability that it is of species \(B\).
(2)
(e)
A biologist decides to classify every crab with \(W\gt c\) as species \(B\). Find the value of \(c\) for which the probability that a crab with \(W\gt c\) is actually of species \(A\) is 0.1.
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(z=1.28155\ldots\) for the upper 10%
(A1)
\(z=-0.674489\ldots\) for the lower 25%
(A1)
\(\mu+1.28155\sigma=60\) and \(\mu-0.674489\sigma=50\)
solving with GDC (graph or table of the expression against \(c\))
(M1)
\(c=63.5\) mm
A1
\(63.4552\ldots\) Accept 63.4–63.5 from rounded parameters.
Answers: (a) \(\mu_{A}=53.4\) mm, \(\sigma_{A}=5.11\) mm · (b) 7.72 mm · (c) 0.279 · (d) 0.749 · (e) \(c=63.5\)
107AAHL4-q00744.13 Bayes’ theoremPaper 2Hard15 marks
Bags of flour labelled 1 kg are filled by three machines, A, B and C, which fill 50%, 30% and 20% of the bags respectively. A bag is underweight if its mass is less than 1000 g.
The masses, in grams, of bags filled by machine A are normally distributed with mean 1010 and standard deviation 6, and those filled by machine C are normally distributed with mean 1012 and standard deviation 5.
(a)
Find the probability that a bag filled by machine A is underweight.
(2)
(b)
The masses of bags filled by machine B are normally distributed with mean 1008 g and standard deviation \(\sigma_{B}\) g, and 2.5% of these bags are underweight. Find \(\sigma_{B}\).
(3)
(c)
Show that the probability that a randomly chosen bag is underweight is 0.0330, correct to three significant figures.
(2)
(d)
A randomly chosen bag is underweight. Find the probability that it was filled by machine A.
(2)
(e)
Machine A is recalibrated: its standard deviation stays at 6 g, but its mean is changed to \(\mu_{A}\) g, so that only 30% of underweight bags come from machine A. Machines B and C are unchanged. Find \(\mu_{A}\), giving your answer correct to one decimal place.
(4)
(f)
After the recalibration, bags from all three machines are mixed and packed in boxes of 20. Find the probability that a box contains at least one underweight bag.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(M\lt 1000)\) with \(M\sim\mathrm{N}(1010,\,6^{2})\)
(M1)
\(0.0478\)
A1
\(0.0477903\ldots\)
(b)
\(\dfrac{1000-1008}{\sigma_{B}}=z\) where \(\mathrm{P}(Z\lt z)=0.025\)
M1
\(z=-1.95996\ldots\)
(A1)
\(\sigma_{B}=4.08\)
A1
\(4.08170\ldots\)
(c)
\(0.5(0.0477903)+0.3(0.025)+0.2(0.00819753)\)
M1
Must see the value \(0.00819753\ldots\) for machine C.
\(0.0330346\ldots\)
A1
\(0.0330\)
AG
(d)
\(\dfrac{0.5(0.0477903)}{0.0330346}\)
(M1)
\(0.723\)
A1
\(0.723336\ldots\)
(e)
\(\dfrac{0.5p}{0.5p+0.3(0.025)+0.2(0.00819753)}=0.3\), where \(p\) is the new probability that a bag from A is underweight
M1
Bayes’ theorem with the new unknown probability.
\(p=0.00783386\ldots\)
A1
From \(0.35p=0.3(0.00913951)\).
\(\dfrac{1000-\mu_{A}}{6}=-2.41657\ldots\)
M1
Standardizing with the \(z\)-value for a lower tail of \(p\).
\(\mu_{A}=1014.5\)
A1
\(1014.4994\ldots\) Accept 1014.4–1014.6.
(f)
\(\mathrm{P}(\text{underweight})=0.5(0.00783386)+0.0091395=0.0130564\), then \(1-(1-0.0130564)^{20}\)
(M1)
Binomial \(\mathrm{B}(20,\,0.0130564)\). FT their \(p\).
108AAHL4-q00754.13 Bayes’ theoremPaper 2Hard13 marks
The level, \(X\) units, of a biomarker in the blood is normally distributed with mean 40 and standard deviation 6 for healthy people, and normally distributed with mean 58 and standard deviation 8 for people who have a particular disease. In a population, 2% of people have the disease. A test is declared positive if \(X\gt c\), where \(c\) is a cut-off value chosen by the laboratory.
(a)
For \(c=50\), find the probability of a positive test for a person with the disease and for a healthy person.
(3)
(b)
For \(c=50\), a randomly chosen person tests positive. Find the probability that this person has the disease.
(3)
(c)
For \(c=50\), a randomly chosen person tests negative. Find the probability that this person has the disease.
(2)
(d)
The laboratory wants 95% of people with the disease to test positive. Find the value of \(c\).
(2)
(e)
For this value of \(c\), find the probability that a person who tests positive has the disease, and comment on your answer in comparison with part (b).
(3)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{P}(X\gt 50)\) for \(\mathrm{N}(58,\,8^{2})\) and for \(\mathrm{N}(40,\,6^{2})\)
\(0.0845664\ldots\) Accept 0.0838–0.0847 (0.0838 from \(c=44.8\)).
lowering the cut-off detects more people with the disease, but many more healthy people now test positive, so a positive result is much less reliable (8.5% compared with 26%)
109AAHL4-q01034.14 Variance, continuous random variables and linear transformationsPaper 2Hard8 marks
A dart lands at a random point on a circular target of radius 5 cm. The target is divided by circles of radius 1 cm and 3 cm, with the same centre, into three regions scoring 10, 5 and 1 points, as shown. The distance, \(X\) cm, of the dart from the centre has probability density function \(f(x)=\dfrac{2x}{25}\) for \(0\le x\le5\), and \(f(x)=0\) otherwise.
Diagram not to scale
(a)
Show that \(\mathrm{P}(X\le a)=\dfrac{a^{2}}{25}\) for \(0\le a\le5\), and explain why this is consistent with the dart being equally likely to land anywhere on the target.
(2)
(b)
Let \(S\) be the score from one dart. Find the probability distribution of \(S\).
(2)
(c)
Find the entry fee per dart that would make the game fair, if a player receives €1 for each point scored.
110AAHL4-q01094.14 Variance, continuous random variables and linear transformationsPaper 2Hard15 marks
Each morning a florist buys \(n\) bouquets from a supplier at €8 each and sells them at €20 each. Bouquets not sold by the end of the day are thrown away. The daily demand, \(D\), for bouquets has the probability distribution shown, and is independent of the number of bouquets bought.
\(d\)
\(2\)
\(3\)
\(4\)
\(5\)
\(\mathrm{P}(D=d)\)
\(0.1\)
\(0.3\)
\(0.4\)
\(0.2\)
The number of bouquets sold on a day is the smaller of \(D\) and \(n\). Let \(G\) be the florist’s daily profit, in euros, from bouquets.
(a)
Find \(\mathrm{E}(D)\) and \(\mathrm{Var}(D)\).
(3)
(b)(i)
For \(n=5\), explain why \(G=20D-40\).
(1)
(b)(ii)
Hence, for \(n=5\), find \(\mathrm{E}(G)\) and the standard deviation of \(G\).
(3)
(c)(i)
For \(n=4\), write down the possible values of \(G\) and find \(\mathrm{E}(G)\).
(3)
(c)(ii)
It is given that \(\mathrm{E}(G)=24\) when \(n=2\) and \(\mathrm{E}(G)=34\) when \(n=3\). State, with a reason, the number of bouquets the florist should buy each morning to maximize the expected profit.
(1)
(d)
For \(n=4\), find the standard deviation of \(G\).
(2)
(e)
The supplier raises the price to €10 per bouquet. For \(n=4\), find the new expected daily profit and the new standard deviation of the daily profit, justifying your answer.
(2)
Show mark scheme
Working
Mark
Notes
(a)
\(\mathrm{E}(D)=0.2+0.9+1.6+1=3.7\)
A1
\(\mathrm{E}(D^{2})=0.4+2.7+6.4+5=14.5\)
(M1)
\(\mathrm{Var}(D)=14.5-3.7^{2}=0.81\)
A1
(b)(i)
\(D\le5\) always, so all \(D\) bouquets demanded are sold: income \(20D\), cost \(5\times8=40\)
R1
(b)(ii)
\(\mathrm{E}(G)=20(3.7)-40=34\) (euros)
A1
FT their (a).
\(\mathrm{Var}(G)=20^{2}\times0.81=324\)
(M1)
Award M0 for \(20\times0.81\) or \(20^{2}\times0.81-40\).
standard deviation \(=18\) (euros)
A1
(c)(i)
\(G=8,\ 28,\ 48\) (for \(D=2,\ 3,\ D\ge4\))
A1
Income \(20\min(D,4)\) minus cost \(32\).
\(\mathrm{E}(G)=8(0.1)+28(0.3)+48(0.6)\)
(M1)
\(\mathrm{P}(G=48)=\mathrm{P}(D\ge4)=0.6\).
\(\mathrm{E}(G)=38\) (euros)
A1
(c)(ii)
\(n=4\), since \(38\gt34\) (for \(n=3\) and \(n=5\)) \(\gt24\)
R1
Must compare all four expected values; \(n\gt5\) cannot help as at most 5 are sold.
111AAHL4-q01104.14 Variance, continuous random variables and linear transformationsPaper 2Hard14 marks
A shop is open from 07:00 to 19:00. The time, \(T\) hours after 07:00, at which a randomly chosen customer enters the shop is modelled by the probability density function
\(h'(a)\gt0\) for \(a\gt1\), so \(h\) is increasing and \(h(a)\gt h(1)=0\); hence mean \(\gt\) median
R1
AG
Answers: (d) \(a=4.92\) · (e) 0.922
113AAHL4-q00324.7 Discrete random variablesPaper 3Hard28 marks
This question investigates the largest score obtained when a fair \(n\)-sided die, with faces numbered \(1,2,\ldots,n\), is rolled several times. The rolls are independent. Let \(M_{r}\) be the largest score obtained when the die is rolled \(r\) times, where \(n,r\in\mathbb{Z}^{+}\) and \(n\ge2\).
(a)(i)
Consider \(n=4\) and \(r=2\).
Show that \(\mathrm{P}(M_{2}=k)=\dfrac{2k-1}{16}\) for \(k=1,2,3,4\).
(2)
(a)(ii)
Find \(\mathrm{E}(M_{2})\).
(2)
(b)(i)
Consider a general value of \(n\), with \(r=2\).
Show that \(\mathrm{P}(M_{2}=k)=\dfrac{2k-1}{n^{2}}\) for \(k=1,2,\ldots,n\).
(2)
(b)(ii)
Given that \(\displaystyle\sum_{k=1}^{n}k^{2}=\dfrac{n(n+1)(2n+1)}{6}\), show that \(\mathrm{E}(M_{2})=\dfrac{(n+1)(4n-1)}{6n}\).
Write down an expression for \(\mathrm{P}(M_{r}\le k)\), for \(k=0,1,\ldots,n\).
(1)
(c)(ii)
Show that \(\mathrm{E}(M_{r})=\displaystyle\sum_{k=0}^{n-1}\mathrm{P}(M_{r}\gt k)\).
(3)
(c)(iii)
Hence show that \(\mathrm{E}(M_{r})=n-\dfrac{1}{n^{r}}\displaystyle\sum_{k=1}^{n-1}k^{r}\).
(2)
(d)
For \(r=3\), use the result in part (c)(iii) to find the value of \(\dfrac{\mathrm{E}(M_{3})}{n}\) when \(n=10\) and when \(n=1000\).
(2)
(e)
Using your answers to parts (b)(iii) and (d), suggest an expression, in terms of \(r\), for \(\displaystyle\lim_{n\to\infty}\dfrac{\mathrm{E}(M_{r})}{n}\).
(1)
(f)
By considering rectangles of width 1 under and over the graph of \(y=x^{r}\), for \(x\ge0\), explain why \(\displaystyle\int_{0}^{n-1}x^{r}\,\mathrm{d}x\lt\sum_{k=1}^{n-1}k^{r}\lt\int_{1}^{n}x^{r}\,\mathrm{d}x\).
(3)
(g)
Hence prove your suggestion in part (e).
(4)
Show mark scheme
Working
Mark
Notes
(a)(i)
\(\mathrm{P}(M_{2}\le k)=\left(\dfrac{k}{4}\right)^{2}\), so \(\mathrm{P}(M_{2}=k)=\dfrac{k^{2}-(k-1)^{2}}{16}\)
M1
Or counting the 2k − 1 ordered pairs with largest value k in a 4 × 4 sample space.
\(\displaystyle\sum_{k=1}^{n-1}k^{r}\) is the total area of \(n-1\) rectangles of width 1 and heights \(1^{r},2^{r},\ldots,(n-1)^{r}\)
R1
placing the rectangle of height \(k^{r}\) on \([k,\,k+1]\): it lies under the curve because \(x^{r}\) is increasing, so the sum is less than \(\displaystyle\int_{1}^{n}x^{r}\,\mathrm{d}x\)
R1
placing it on \([k-1,\,k]\): the curve lies under the rectangle, so the sum is greater than \(\displaystyle\int_{0}^{n-1}x^{r}\,\mathrm{d}x\)
114AAHL4-q00774.13 Bayes’ theoremPaper 3Hard29 marks
This question investigates how the probability of the next toss being a head changes as more heads are observed.
A box contains three coins. Coin \(A\) is fair; coin \(B\) has a head on both sides; coin \(C\) is biased so that the probability of a head is \(\tfrac{1}{4}\). One coin is chosen at random from the box and is then tossed repeatedly. For the chosen coin, the tosses are independent.
(a)(i)
The chosen coin is tossed once.
Find the probability that it shows a head.
(2)
(a)(ii)
Given that it shows a head, find the probability that the coin is \(B\).
(2)
(b)
The chosen coin is tossed twice and shows two heads. Find the probability that the coin is \(B\).
(3)
(c)
Show that, given that the first \(n\) tosses are all heads, the probability that the coin is \(B\) is \(\dfrac{4^{n}}{4^{n}+2^{n}+1}\).
(3)
(d)
Find the least value of \(n\) for which this probability exceeds 0.999.
(3)
(e)(i)
Let \(q_{n}\) be the probability that toss \(n+1\) is a head, given that the first \(n\) tosses are all heads, for \(n\ge1\).
Show that \(q_{n}=\dfrac{4^{n+1}+2^{n+1}+1}{4\left(4^{n}+2^{n}+1\right)}\).
(3)
(e)(ii)
Find \(q_{1}\) and \(q_{2}\).
(2)
(e)(iii)
Write down \(\displaystyle\lim_{n\to\infty}q_{n}\).
(1)
(f)
By writing \(t=2^{n}\) and considering \(1-q_{n}\), prove that \(q_{n}\) increases as \(n\) increases.
(4)
(g)(i)
The investigation is generalized. A box contains two coins only, with probabilities of a head \(p\) and \(s\), where \(0\lt p\lt s\le1\). One coin is chosen at random and tossed repeatedly. Let \(q_{n}\) now be the probability that toss \(n+1\) is a head, given that the first \(n\) tosses are all heads, with \(q_{0}\) the probability that the first toss is a head.
Show that \(q_{n}=\dfrac{p^{n+1}+s^{n+1}}{p^{n}+s^{n}}\).
(2)
(g)(ii)
Prove that \(q_{n}\gt q_{n-1}\) for all \(n\in\mathbb{Z}^{+}\).
Show that the mode of \(X_{n}\) is \(\dfrac{n}{n+1}\).
(2)
(c)(ii)
Show that \(\mathrm{E}(X_{n})=\dfrac{n+1}{n+3}\).
(3)
(c)(iii)
Prove that the mode of \(X_{n}\) is greater than or equal to \(\mathrm{E}(X_{n})\), with equality only when \(n=1\).
(3)
(d)(i)
Show that \(\mathrm{E}(X_{n}^{2})=\dfrac{(n+1)(n+2)}{(n+3)(n+4)}\).
(2)
(d)(ii)
Hence show that \(\mathrm{Var}(X_{n})=\dfrac{2(n+1)}{(n+3)^{2}(n+4)}\).
(3)
(d)(iii)
Find \(\displaystyle\lim_{n\to\infty}\mathrm{Var}(X_{n})\) and interpret this result.
(2)
(e)(i)
Let \(I(p,q)=\displaystyle\int_{0}^{1}x^{p}(1-x)^{q}\,\mathrm{d}x\), where \(p,q\in\mathbb{N}\).
For \(q\ge1\), use integration by parts to show that \(I(p,q)=\dfrac{q}{p+1}\,I(p+1,\,q-1)\).
(3)
(e)(ii)
Hence prove that \(I(p,q)=\dfrac{p!\,q!}{(p+q+1)!}\).
(4)
(e)(iii)
Write down, in terms of \(p\) and \(q\), the value of the constant \(c\) for which \(g(x)=c\,x^{p}(1-x)^{q}\), \(0\le x\le1\), is a probability density function.
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