Edexcel International A Level Maths · S2 · Statistics 2
Binomial and Poisson distributions
28 free, high-quality past-paper-style questions on Binomial and Poisson distributions for Edexcel IAL Maths, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Maths notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 236 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked
1IALM-S2-1-B-01S2 · A2Easy7 marks
A darts player aims at the bullseye. On each throw the probability that she hits it is 0.2, independently of all other throws. She makes 10 throws and X is the number of times she hits the bullseye.
(a)
State the distribution of X, and give two assumptions, in context, that are needed for it.
(2)
(b)
Find the probability that she hits the bullseye exactly 3 times.
(2)
(c)
Find the probability that she hits the bullseye at most twice.
(1)
(d)
Find the probability that she hits the bullseye at least 4 times.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
X ~ B(10, 0.2)
B1
any two in context, e.g. the throws are independent of each other; the probability of hitting the bullseye is 0.2 on every throw
B1
Part (b)
P(X = 3) = 10C3(0.2)3(0.8)7
M1
= 0.2013
A1
Part (c)
P(X ≤ 2) = 0.6778
B1
Part (d)
1 − P(X ≤ 3)
M1
= 1 − 0.8791 = 0.1209
A1
Common error: Writing P(X ≥ 4) = 1 − P(X ≤ 4). The complement of 'at least 4' is 'at most 3'.
Seeds of a particular plant are sown 5 to a pot, and X is the number of seeds in a pot that germinate, where X ~ B(5, p). Figure 1 shows the probability distribution of X, with each probability given to 4 decimal places. The bar for x = 3 has been left out.
(a)
State one assumption, in context, that is needed for the binomial model.
(1)
(b)
Use Figure 1 to find P(X = 3).
(2)
(c)
Use the probability for x = 0 to show that p = 0.4.
(2)
(d)
Find the mean and the variance of X.
(2)
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Answer
Mark
Notes
Part (a)
e.g. the seeds germinate independently of each other, or each seed has the same probability of germinating
B1
Part (b)
1 − (0.0778 + 0.2592 + 0.3456 + 0.0768 + 0.0102)
M1
= 1 − 0.7696 = 0.2304
A1
Part (c)
(1 − p)5 = 0.0778
M1
1 − p = 0.07781/5 = 0.600 to 3 s.f., so p = 0.4
A1
Part (d)
mean = 5 × 0.4 = 2
B1
variance = 5 × 0.4 × 0.6 = 1.2
B1
Exam tip: The probabilities of a distribution add up to 1, so one missing value can always be found by subtraction.
Dandelions grow at random in a park, at a mean rate of 0.05 per square metre. Figure 1 shows two regions of the park: region A is a rectangle and region B is a right-angled triangle, with the dimensions shown.
(a)
Find the probability that region A contains exactly 3 dandelions.
(2)
(b)
Find the probability that region B contains at most 1 dandelion.
(2)
(c)
Find the probability that regions A and B together contain more than 6 dandelions.
(2)
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Answer
Mark
Notes
Part (a)
area 90 m2, so Po(4.5)
M1
e−4.5 4.53/3! = 0.1687
A1
Part (b)
area ½ × 8 × 5 = 20 m2, so mean λ = 0.05 × 20 = 1.0
B1
P(≤ 1) = 0.7358
A1
Part (c)
total ~ Po(5.5) and 1 − P(≤ 6)
M1
= 1 − 0.6860 = 0.3140
A1
Common error: Using the full base × height for a triangle. Region B has area ½ × 8 × 5 = 20 m², so its mean is 1, not 2.
A gardener counts the aphids on each of 60 leaves picked at random from a rose bush. The bar chart in Figure 1 shows the number of leaves with each number of aphids, x.
(a)
Calculate the mean and the variance of the number of aphids per leaf.
(3)
(b)
Explain whether your answers support a Poisson model for the number of aphids on a leaf.
(1)
(c)
Suggest a reason, in context, why a Poisson model may fail here.
(1)
(d)
Explain why a binomial distribution would not be a suitable model either.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
Σfx = 120, mean = 120/60 = 2
B1
Σfx2 = 492, variance = 492/60 − 22
M1
= 4.2 (accept 4.27 using the divisor n − 1)
A1
Part (b)
no: the variance is about twice the mean, but a Poisson distribution has mean = variance
B1
Part (c)
e.g. aphids live in colonies, so they cluster on some leaves and are not independent of each other
B1
Part (d)
a binomial variance np(1 − p) is less than its mean np, but here the variance is greater than the mean
B1
Exam tip: Compare mean and variance: equal suggests Poisson, variance mean suggests clustering.
A courier has 9 parcels to deliver. For each parcel, independently, the probability that nobody is in to receive it is 0.35. X is the number of parcels for which nobody is in. Figure 1 shows the graph of F(x) = P(X ≤ x), with the height of each step given to 4 decimal places.
(a)
Use Figure 1 to find P(X = 3).
(2)
(b)
Find P(2 < X ≤ 6).
(2)
(c)
Find P(X ≥ 5).
(1)
(d)
State the most likely value of X, giving a reason based on Figure 1.
(1)
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Answer
Mark
Notes
Part (a)
F(3) − F(2)
M1
= 0.6089 − 0.3373 = 0.2716
A1
Part (b)
F(6) − F(2)
M1
= 0.9888 − 0.3373 = 0.6515
A1
Part (c)
1 − F(4) = 1 − 0.8283 = 0.1717
B1
Part (d)
3: the jump at x = 3 (0.2716) is the largest step in the graph
B1
Exam tip: The height of the jump at x is P(X = x). A cumulative graph turns every 'exactly' into a subtraction of two step heights.
A courier company finds that 1.5% of parcels are damaged in transit, independently of each other. A van carries 200 parcels and X is the number of damaged parcels in the van.
(a)
Explain why the distribution of X can be approximated by a Poisson distribution, and state its parameter.
(2)
(b)
Use the approximation to find the probability that exactly 4 parcels are damaged.
(2)
(c)
Use the approximation to find the probability that at most 1 parcel is damaged.
(2)
(d)
Calculate the exact binomial value of P(X ≤ 1) and comment on the approximation.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
X ~ B(200, 0.015), with n large and p small
B1
X ≈ Po(3), since np = 3
B1
Part (b)
e−3 34/4!
M1
= 0.1680
A1
Part (c)
P(X ≤ 1) from the Po(3) table, or e−3(1 + 3)
M1
= 0.1991
A1
Part (d)
0.985200 + 200(0.015)(0.985)199 = 0.1969
B1
the approximation differs by about 0.002, so it is good
B1
Common error: Using the Poisson approximation without saying why. State both conditions: n is large and p is small.
A wildlife camera at a pond records visits by herons at a mean rate of 1.2 per day and visits by kingfishers at a mean rate of 0.8 per day. Visits by each species occur at random.
(a)
Find the probability that no birds of either species visit on a given day.
(2)
(b)
Find the probability that there are at least 5 visits in total over a 2-day period.
(3)
(c)
Find the probability that on a given day there is exactly 1 heron visit and exactly 1 kingfisher visit.
(2)
(d)
State the assumption about the two species that you have used in parts (a) to (c).
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
total ~ Po(2)
M1
P(0) = e−2 = 0.1353
A1
Part (b)
Po(4) for 2 days
B1
1 − P(≤ 4)
M1
= 1 − 0.6288 = 0.3712
A1
Part (c)
1.2e−1.2 × 0.8e−0.8
M1
= 0.96e−2 = 0.1299
A1
Part (d)
the numbers of heron visits and kingfisher visits are independent of each other
B1
Exam tip: 'Exactly 1 of each' is not the same as 'exactly 2 in total': multiply the two separate Poisson probabilities.
An ecologist counts the number of weeds, x, in each of 80 one-metre quadrats on a lawn. The results are: 0 weeds in 18 quadrats, 1 weed in 26, 2 weeds in 20, 3 weeds in 10, 4 weeds in 4 and 5 weeds in 2. (You may use Σfx = 122 and Σfx2 = 310.)
(a)
Calculate the mean and the variance of the number of weeds per quadrat.
(3)
(b)
Explain how your answers support a Poisson model for the number of weeds in a quadrat.
(1)
(c)
Using a Poisson distribution with λ = 1.5, find the expected number of the 80 quadrats with no weeds.
(2)
(d)
Using the same model, find the probability that a quadrat contains more than 3 weeds.
(2)
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Answer
Mark
Notes
Part (a)
mean = 122/80 = 1.525
B1
variance = 310/80 − 1.5252
M1
= 1.549 (accept 1.55, or 1.569 using the divisor n − 1)
A1
Part (b)
the mean and variance are close, and a Poisson distribution has mean equal to variance
B1
Part (c)
80 × e−1.5
M1
= 17.85 (compared with 18 observed)
A1
Part (d)
1 − P(≤ 3)
M1
= 1 − 0.9344 = 0.0656
A1
Exam tip: Mean ≈ variance supports a Poisson model; a variance much larger than the mean suggests clustering, so a Poisson model would be doubtful.
The weekly sales of one model of camera in a shop are modelled by a Poisson distribution with mean 2.5.
(a)
Give two conditions, in context, for the weekly number of sales to follow a Poisson distribution.
(2)
(b)
Find the probability that exactly 3 cameras are sold in a given week.
(2)
(c)
The shop has s cameras in stock at the start of each week and is not restocked during the week. Find the smallest value of s for which the probability that the shop cannot meet the demand is less than 0.05.
(3)
(d)
State the standard deviation of the number of cameras sold in a week.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
sales occur independently of each other
B1
sales occur one at a time, at a constant mean rate
B1
Part (b)
e−2.5 2.53/3!
M1
= 0.2138
A1
Part (c)
need P(X > s) < 0.05, i.e. P(X ≤ s) > 0.95
M1
P(X ≤ 4) = 0.8912, P(X ≤ 5) = 0.9580
M1
s = 5
A1
Part (d)
√2.5 = 1.581
B1
Common error: Giving the variance (2.5) when the standard deviation is asked for. For a Poisson distribution the standard deviation is √λ.
Light bulbs are sold in boxes of 50. Each bulb, independently, is faulty with probability 0.04, and X is the number of faulty bulbs in a box. Figure 1 compares the distribution of X with the Poisson distribution Po(2); its table gives both sets of probabilities to 4 decimal places.
(a)
Explain why Po(2) is a suitable approximation for the distribution of X.
(2)
(b)
Use the table in Figure 1 to find P(X ≤ 2) using each distribution, and comment on your answers.
(3)
(c)
Find the percentage error in P(X = 0) when the approximation is used.
(2)
(d)
By comparing the variances of the two distributions, explain why the Poisson bars in Figure 1 are taller than the binomial bars for x ≥ 5.
(1)
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Answer
Mark
Notes
Part (a)
X ~ B(50, 0.04) with n large and p small
B1
mean np = 50 × 0.04 = 2
B1
Part (b)
binomial: 0.1299 + 0.2706 + 0.2762 = 0.6767
B1
Poisson: 0.1353 + 0.2707 + 0.2707 = 0.6767
B1
they agree to 4 d.p., so the approximation is very good for this probability
B1
Part (c)
(0.1353 − 0.1299)/0.1299 × 100
M1
= 4.16% (allow 4.1% to 4.2%)
A1
Part (d)
Var(X) = 50 × 0.04 × 0.96 = 1.92 < 2, so the Poisson distribution is more spread out and puts more probability in the upper tail
B1
Exam tip: A Poisson approximation keeps the mean but slightly overstates the variance (λ instead of np(1 − p)), so its tails are a little heavier.
A factory receives large batches of washers. From each batch a random sample of 10 washers is tested, and the batch is accepted if at most 1 of them is defective. Figure 1 shows the probability that a batch is accepted, plotted against p, the proportion of defective washers in the batch.
(a)
Give a reason why the number of defective washers in the sample can be modelled by a binomial distribution, even though the washers are not replaced.
(1)
(b)
Calculate the probability that a batch with p = 0.1 is accepted.
(2)
(c)
Use Figure 1 to estimate the value of p for which a batch is equally likely to be accepted or rejected.
(1)
(d)
A supplier sends 5 batches, each with p = 0.25. Find the probability that at least one of the batches is accepted.
(4)
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Answer
Mark
Notes
Part (a)
the batch is large, so taking 10 washers hardly changes the proportion defective: the trials are almost independent with constant p
B1
Part (b)
B(10, 0.1) and P(X ≤ 1)
M1
= 0.7361
A1
Part (c)
reading across from 0.5: p ≈ 0.16 (allow 0.15 to 0.17)
B1
Part (d)
P(accepted) = P(X ≤ 1) for B(10, 0.25)
M1
= 0.2440
A1
1 − (1 − 0.2440)5
M1
= 0.7531
A1
Exam tip: When a probability from one model is used again, as the 'p' of 5 independent batches, write the new distribution down before calculating.
The number of calls to a helpline in a 10-minute period is modelled by a Poisson distribution with mean λ. Figure 1 shows this distribution, with each probability given to 4 decimal places.
(a)
Use the probabilities in Figure 1 for x = 0 and x = 1 to show that λ = 3.5.
(2)
(b)
Find the probability that more than 5 calls are received in a 10-minute period.
(2)
(c)
Find the probability that at least 10 calls are received in a 20-minute period.
(2)
(d)
State the mean and the variance of the number of calls received in one hour.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
P(X = 1) ÷ P(X = 0) = λe−λ ÷ e−λ = λ
M1
λ = 0.1057 ÷ 0.0302 = 3.5
A1
Part (b)
1 − P(X ≤ 5)
M1
= 1 − 0.8576 = 0.1424 (allow 0.1423 from adding the values in Figure 1)
A1
Part (c)
Po(7) and 1 − P(≤ 9)
M1
= 1 − 0.8305 = 0.1695
A1
Part (d)
Po(21): mean 21 and variance 21
B1
Exam tip: For a Poisson distribution, P(X = r + 1) ÷ P(X = r) = λ/(r + 1). Two neighbouring probabilities are enough to find λ.
An alarm unit contains three components A, B and C, connected between terminals P and Q as shown in Figure 1. The unit works if there is a path from P to Q through working components only. The components work independently, with the probabilities shown.
(a)
Show that the probability that a unit works is 0.8.
(3)
(b)
A school fits 20 of these units, which work independently. Find the probability that at most 2 of the units fail.
(2)
(c)
Find the probability that more than 16 of the 20 units work.
(2)
(d)
Find the mean and the standard deviation of the number of the 20 units that work.
(2)
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Answer
Mark
Notes
Part (a)
upper path works: 0.75 × 0.8 = 0.6
M1
unit fails only if both paths fail: (1 − 0.6)(1 − 0.5) = 0.2
Customers arrive at a café at random. Figure 1 shows how the mean rate of arrivals, in customers per hour, changes during the day; within each period shown the rate is constant. Arrivals in non-overlapping time intervals are independent.
(a)
Find the probability that exactly 2 customers arrive between 09:00 and 09:30.
(2)
(b)
Find the probability that at least 4 customers arrive between 08:30 and 09:30.
(3)
(c)
Find the probability that fewer than 6 customers arrive between 10:30 and 11:30.
(2)
(d)
Find the variance of the number of customers who arrive between 07:00 and 15:00.
(1)
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Answer
Mark
Notes
Part (a)
rate 2 per hour for ½ hour: mean λ = 1.0
M1
e−1 12/2! = 0.1839
A1
Part (b)
08:30–09:00 ~ Po(6 × 0.5) = Po(3) and 09:00–09:30 ~ Po(2 × 0.5) = Po(1.0); these are independent, so the total ~ Po(3 + 1.0) = Po(4)
M1
1 − P(≤ 3)
M1
= 1 − 0.4335 = 0.5665
A1
Part (c)
independent Po(2 × 0.5) = Po(1.0) and Po(10 × 0.5) = Po(5), so the total ~ Po(1.0 + 5) = Po(6)
M1
P(≤ 5) = 0.4457
A1
Part (d)
the four periods give independent Po(12), Po(4), Po(20) and Po(8), so the total ~ Po(44) and variance = mean = 44
B1
Exam tip: When an interval crosses a change of rate, find the mean for each piece and add them: the counts in the pieces are independent Poisson variables.
A rare blood type occurs in 0.8% of a population. A clinic tests 250 donors, chosen at random, and X is the number with the rare type.
(a)
Find the mean and the variance of X.
(2)
(b)
Explain why a Poisson distribution is a suitable approximation for X, referring to your answers to part (a).
(2)
(c)
Use a Poisson approximation to find P(X ≥ 4).
(2)
(d)
Using a Poisson approximation, find the least number of donors, n, that must be tested for the probability of finding at least one donor with the rare type to exceed 0.95.
(3)
(e)
Repeat the calculation in part (d) using the exact binomial distribution, and comment on the result.
(2)
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Answer
Mark
Notes
Part (a)
X ~ B(250, 0.008); mean = 250 × 0.008 = 2
B1
variance = 2 × 0.992 = 1.984
B1
Part (b)
n = 250 is large and p = 0.008 is small
B1
the mean and variance are almost equal, as they are for a Poisson distribution, so Po(2) is suitable
B1
Part (c)
1 − P(≤ 3) with Po(2)
M1
= 1 − 0.8571 = 0.1429
A1
Part (d)
Po(0.008n): 1 − e−0.008n > 0.95
M1
n > ln 20 / 0.008 = 374.5
M1
n = 375
A1
Part (e)
1 − 0.992n > 0.95 gives n > ln 0.05 / ln 0.992 = 372.97, so n = 373
B1
very close to 375: the approximation is good here, and slightly cautious
B1
Exam tip: If asked to 'explain' an approximation using earlier answers, quote them: mean 2 and variance 1.984 are almost equal.
The meteors seen by an observer on a clear night are modelled as a Poisson process with an average of 9 sightings per hour.
(a)
Find the probability that she sees fewer than 5 meteors in an hour.
(2)
(b)
Find the probability that she sees exactly 2 meteors in a 10-minute period.
(2)
(c)
In one hour she sees exactly 7 meteors. Find the probability that exactly 1 of them was seen in the first 20 minutes.
(4)
(d)
Find the length of time, in minutes, for which the probability that she sees no meteors is 0.5.
(3)
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Answer
Mark
Notes
Part (a)
X ~ Po(9), P(X ≤ 4)
M1
= 0.0550
A1
Part (b)
Po(1.5): e−1.5 1.52/2!
M1
= 0.2510
A1
Part (c)
requires 1 meteor in the first 20 minutes (Po(3)) and 6 in the last 40 minutes (Po(6)), independent intervals
M1
P = [3e−3 × e−666/6!] ÷ [e−997/7!]
M1
the exponentials cancel: 7 × 3 × 66/97
A1
= 448/2187 = 0.2048
A1
Part (d)
in t minutes the rate is 0.15t: e−0.15t = 0.5
M1
t = ln 2 / 0.15
M1
= 4.62 minutes
A1
Exam tip: Events in non-overlapping time intervals of a Poisson process are independent, so a conditional probability splits into a product over the separate intervals.
The probability that a visitor to an online shop makes a purchase is 0.04, independently of other visitors. In a particular hour, 125 visitors use the shop and X is the number of purchases.
(a)
State the exact distribution of X and find its mean and variance.
(2)
(b)
Using a suitable approximation, find the probability that there are at least 8 purchases in the hour.
(3)
(c)
Assuming the same number of visitors in each hour and that hours are independent, find the probability that there are at least 8 purchases in each of 3 given hours.
(2)
(d)
Use the approximation to find the probability that the number of purchases in an hour is within one standard deviation of the mean.
(3)
(e)
Comment on the variance of the approximating distribution compared with the exact variance.
(1)
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Answer
Mark
Notes
Part (a)
X ~ B(125, 0.04)
B1
mean 5, variance 4.8
B1
Part (b)
n large, p small, so X ≈ Po(5)
B1
1 − P(X ≤ 7)
M1
= 1 − 0.8666 = 0.1334
A1
Part (c)
0.13343
M1
= 0.002374 (accept 0.00237 from the unrounded value)
A1
Part (d)
standard deviation √5 = 2.236…, so 2.76 < X < 7.24
M1
P(X ≤ 7) − P(X ≤ 2)
M1
= 0.8666 − 0.1247 = 0.7419
A1
Part (e)
the Poisson variance 5 is slightly larger than the exact 4.8, but close because p is small
B1
Exam tip: For 'within one standard deviation', work out the interval as decimals first, then list the whole numbers inside it before turning to the tables.
A firm makes plastic connectors on two machines, A and B. Figure 1 shows the share of the output made on each machine and the percentage of each machine's connectors that are defective. All connectors go into one mixed store, and crates of 250 connectors are filled at random from the store.
(a)
Show that the probability that a connector chosen at random from the store is defective is 0.032.
(2)
(b)
X is the number of defective connectors in a crate. Explain why a Poisson distribution is a suitable approximation for X, and state its mean.
(2)
(c)
Use the approximation to find the probability that a crate contains more than 10 defective connectors.
(2)
(d)
A customer orders a crate of 250 connectors made only on machine A and a crate of 100 connectors made only on machine B. Using Poisson approximations, find the probability that the two crates together contain at most 6 defective connectors.
(3)
(e)
A connector chosen at random from the mixed store is defective. Find the probability that it was made on machine B.
(2)
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Answer
Mark
Notes
Part (a)
0.6 × 0.02 + 0.4 × 0.05
M1
= 0.012 + 0.020 = 0.032
A1
Part (b)
X ~ B(250, 0.032), with n large and p small
B1
X ≈ Po(8), since 250 × 0.032 = 8
B1
Part (c)
1 − P(≤ 10)
M1
= 1 − 0.8159 = 0.1841
A1
Part (d)
A: 250 × 0.02 = 5, B: 100 × 0.05 = 5, so Po(5) and Po(5)
M1
independent, so the total ~ Po(10)
M1
P(≤ 6) = 0.1301
A1
Part (e)
0.4 × 0.05 ÷ 0.032
M1
= 0.625
A1
Exam tip: Find the overall probability of a defect first; only then does each crate become a single binomial, B(250, 0.032), that can be approximated.
Figure 1 shows the times at which a taxi firm received calls between 08:00 and 12:00 on one morning. The dashed lines divide the period into 16 intervals of 15 minutes.
(a)
Use Figure 1 to find the number of 15-minute intervals in which 0, 1, 2, 3, 4 and 5 calls were received.
(2)
(b)
Find the mean and the variance of the number of calls per 15-minute interval, and explain how your answers support a Poisson model.
(4)
(c)
Using a Poisson distribution with mean 2 per 15 minutes, find the probability that more than 4 calls are received in a 15-minute interval.
(2)
(d)
Find the probability that at most 5 calls are received in one hour.
(2)
(e)
In a particular half-hour exactly 4 calls are received. Find the probability that 2 of them were received in each quarter-hour.
the mean and the variance are close, as they are equal for a Poisson distribution
B1
Part (c)
1 − P(≤ 4) for Po(2)
M1
= 1 − 0.9473 = 0.0527
A1
Part (d)
Po(8)
M1
P(≤ 5) = 0.1912
A1
Part (e)
P(2 in first) × P(2 in second) ÷ P(4 in Po(4))
M1
= 4C2(½)4 = 0.375
A1
Common error: Counting a call twice or missing one when reading a timeline. Check that your frequencies add up to 16 intervals and that Σfx equals the total number of dots.
Figure 1 shows the graph of P(X ≤ 3) against λ, where X ~ Po(λ). A garage keeps 3 spare wiper motors and is restocked every d days. Demand for the motors occurs at random, at a mean rate of 0.25 per day. The garage wants the probability that the demand before restocking is at most 3 to be at least 0.9.
(a)
Use Figure 1 to estimate P(X ≤ 3) when λ = 4, and then find its value from tables.
(2)
(b)
Use Figure 1 to estimate the largest value of λ for which P(X ≤ 3) ≥ 0.9.
(1)
(c)
Find the largest whole number d that meets the garage's requirement, showing calculations to justify your answer.
(3)
(d)
With d = 6, find the probability that at least 2 motors are demanded before restocking.
(2)
(e)
Demand for brake pads at the garage occurs at random, independently of the motors, at a mean rate of 0.15 per day. Find the probability that the total demand for motors and brake pads over 10 days is at least 6.
(2)
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Answer
Mark
Notes
Part (a)
from the graph about 0.43 (allow 0.42 to 0.45)
B1
tables: 0.4335
B1
Part (b)
where the curve crosses the dashed line: λ ≈ 1.74 (allow 1.7 to 1.8)
B1
Part (c)
λ = 0.25d ≤ about 1.74, so d is about 7 or less
M1
d = 6: Po(1.5), from tables P(≤ 3) = 0.9344 ≥ 0.9; d = 7: Po(1.75) is not in the tables, so P(≤ 3) = e−1.75(1 + 1.75 + 1.752/2! + 1.753/3!) = 0.8992 < 0.9
M1
d = 6
A1
Part (d)
1 − P(≤ 1) for Po(1.5)
M1
= 1 − 0.5578 = 0.4422
A1
Part (e)
motors ~ Po(0.25 × 10) = Po(2.5) and brake pads ~ Po(0.15 × 10) = Po(1.5), independent, so the total ~ Po(2.5 + 1.5) = Po(4)
M1
1 − P(≤ 5) = 1 − 0.7851 = 0.2149
A1
Exam tip: A value read from a graph is only an estimate. When the reading is close to a boundary, calculate the probabilities on both sides of it.
Figure 1 shows a game board. A ball released at the top hits one pin in each row. At each pin, independently, it moves to the right with probability 0.4 and to the left with probability 0.6. The slot it lands in is numbered by the number of times it moved to the right, and X is this slot number. A player wins a prize if the ball lands in a slot marked prize.
(a)
Use Figure 1 to state the distribution of X, explaining why it is suitable.
(2)
(b)
Find the probability that the ball lands in slot 3.
(2)
(c)
Show that the probability that a player wins a prize is 0.1149.
(3)
(d)
Twenty balls are released. Find the probability that at least 3 of them win prizes.
(3)
(e)
Find the mean and the standard deviation of X.
(2)
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Answer
Mark
Notes
Part (a)
there are 8 rows, so 8 independent left/right trials, each with P(right) = 0.4
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