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Edexcel International A Level Maths · S1 · Statistics 1

Discrete random variables

28 free, high-quality past-paper-style questions on Discrete random variables 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 · 256 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · How questions are checked

1IALM-S1-5-B-01 S1 · ASEasy6 marks

The number of text messages, X, that a phone receives in a randomly chosen minute has the probability distribution below, where a is a constant.

x0123
P(X = x)0.350.4a0.05

(a)

Find the value of a.

(1)
(b)

Find P(X ≥ 1).

(1)
(c)

Find E(X).

(2)
(d)

Find Var(X).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
0.35 + 0.4 + a + 0.05 = 1, so a = 0.2B1
Part (b)
1 − 0.35 = 0.65B1
Part (c)
0(0.35) + 1(0.4) + 2(0.2) + 3(0.05)M1
0.95A1
Part (d)
E(X2) = 0.4 + 4(0.2) + 9(0.05) = 1.65 and uses E(X2) − [E(X)]2M1
1.65 − 0.952 = 0.7475A1

Common error: Giving E(X2) = 1.65 as the variance. Subtract the square of the mean: 1.65 − 0.9025.

2IALM-S1-5-B-02 S1 · ASEasy5 marks

A restaurant takes table bookings for groups of 1, 2, 3 or 6 people. The size of a randomly chosen booking is modelled by the random variable X with probability function P(X = x) = k/x for x = 1, 2, 3, 6, where k is a constant.

(a)

Show that k = 1/2.

(2)
(b)

Find P(X ≤ 3).

(1)
(c)

Find E(X).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
k + k/2 + k/3 + k/6 = 1M1
2k = 1, so k = 1/2, as requiredA1
Part (b)
1 − P(X = 6) = 1 − 1/12 = 11/12B1
Part (c)
Σ x × (k/x) = 4k, or 1(1/2) + 2(1/4) + 3(1/6) + 6(1/12)M1
2A1

Exam tip: In a "show that" part, write the equation from Σ P(X = x) = 1 in full before simplifying; the given answer earns nothing without that line.

3IALM-S1-5-B-03 S1 · ASEasy7 marks

The number of strokes, X, that a club golfer takes on a short hole has the cumulative distribution function F(x) shown below.

x12345
F(x)0.150.350.60.851

(a)

Find P(X = 3).

(1)
(b)

Find P(2 ≤ X ≤ 4).

(2)
(c)

Find the probability distribution of X.

(2)
(d)

Given that the golfer takes at least 2 strokes, find the probability that they take more than 3 strokes.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
0.6 − 0.35 = 0.25B1
Part (b)
P(X ≤ 4) − P(X ≤ 1)M1
0.85 − 0.15 = 0.7A1
Part (c)
differences of consecutive values of FM1
0.15, 0.2, 0.25, 0.25, 0.15 for x = 1, 2, 3, 4, 5A1
Part (d)
P(X > 3) / P(X ≥ 2) = 0.4/0.85M1
8/17 or 0.471 (3 s.f.)A1

Exam tip: For whole-number values, P(2 ≤ X ≤ 4) = F(4) − F(1), not F(4) − F(2): the lower end must still be included.

4IALM-S1-5-B-04 S1 · ASEasy6 marks

A computer program picks a whole number X at random from 1, 2, 3, …, 8, each number being equally likely.

(a)

Name the distribution of X and write down P(X = x).

(1)
(b)

Find E(X) and Var(X).

(3)
(c)

The program then outputs S = 2X. Find E(S) and Var(S).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
discrete uniform distribution with P(X = x) = 1/8 for x = 1, 2, …, 8B1
Part (b)
E(X) = (8 + 1)/2 = 4.5B1
uses (82 − 1)/12 or E(X2) − [E(X)]2 with E(X2) = 25.5M1
Var(X) = 63/12 = 5.25A1
Part (c)
E(S) = 2 × 4.5 = 9 ftB1
Var(S) = 22 × 5.25 = 21 ftB1

Common error: Doubling the variance when the variable is doubled. Var(2X) = 4 Var(X), because every squared distance from the mean is multiplied by 4.

5IALM-S1-5-FB-01 S1 · ASEasy7 marks

The number of goals, X, scored by a hockey team in a randomly chosen match is modelled by the probability distribution shown in Figure 1. Each probability lies on a gridline.

00.050.10.150.20.250.30.3501234xP(X = x)
(a)

Write down the most likely number of goals in a match.

(1)
(b)

Find P(X ≥ 2).

(1)
(c)

Find E(X).

(2)
(d)

Find Var(X).

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
the tallest line is at x = 1, so 1 goalB1
Part (b)
0.25 + 0.2 + 0.1 = 0.55B1
Part (c)
0(0.15) + 1(0.3) + 2(0.25) + 3(0.2) + 4(0.1)M1
1.8A1
Part (d)
E(X2) = 1(0.3) + 4(0.25) + 9(0.2) + 16(0.1)M1
= 4.7A1
Var(X) = 4.7 − 1.82 = 1.46A1

Exam tip: Read every height off the chart and check that they add to 1 before you start: 0.15 + 0.3 + 0.25 + 0.2 + 0.1 = 1. A misread height shows up straight away.

6IALM-S1-5-FB-02 S1 · ASEasy7 marks

A commuter travels by train on 4 days each week. The number of days, X, in a randomly chosen week on which the train is late has the cumulative distribution function F(x) shown in Figure 1. The value of F at each step lies on a gridline.

00.10.20.30.40.50.60.70.80.91.0012345xF(x)
(a)

Find P(X = 2).

(1)
(b)

Find P(1 < X ≤ 3).

(2)
(c)

Find the probability distribution of X.

(2)
(d)

Find E(X).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
P(X ≤ 2) − P(X ≤ 1) = 0.65 − 0.3 = 0.35B1
Part (b)
P(X ≤ 3) − P(X ≤ 1)M1
0.9 − 0.3 = 0.6A1
Part (c)
the jumps in F at each value of xM1
P(X = x) = 0.1, 0.2, 0.35, 0.25, 0.1 for x = 0, 1, 2, 3, 4A1
Part (d)
0(0.1) + 1(0.2) + 2(0.35) + 3(0.25) + 4(0.1)M1
2.05 ftA1

Exam tip: On a step graph the probability of a single value is the height of the jump at that value. The filled dot shows the value F takes at the jump; the open dot is the level just before it.

7IALM-S1-5-FB-03 S1 · ASEasy7 marks

Figure 1 shows a spinner at a school fair. When it is spun, the probability that it stops in a sector is proportional to the angle of that sector at the centre. The score, X, is the number shown in the sector where it stops.

1120°290°390°560°Large numbers:scores.Small labelswith °: anglesat the centre.
(a)

Find the probability distribution of X.

(2)
(b)

Find E(X).

(2)
(c)

Find Var(X).

(2)
(d)

Find the probability that a single spin gives a score greater than E(X).

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
angle ÷ 360 for each sectorM1
P(X = 1) = 1/3, P(X = 2) = 1/4, P(X = 3) = 1/4, P(X = 5) = 1/6A1
Part (b)
1(1/3) + 2(1/4) + 3(1/4) + 5(1/6)M1
= 29/12 (≈ 2.42)A1
Part (c)
E(X2) = 1/3 + 4/4 + 9/4 + 25/6 = 31/4, then E(X2) − [E(X)]2M1
31/4 − (29/12)2 = 275/144 (≈ 1.91)A1
Part (d)
X > 2.42 means X = 3 or 5: 1/4 + 1/6 = 5/12 ftB1

Common error: Using 1/4 for every sector because there are four of them. The sectors are not equal, so each probability is its angle divided by 360°.

8IALM-S1-5-FB-04 S1 · ASEasy6 marks

The six cards in Figure 1 are shuffled and placed face down. One card is chosen at random and X is the number on it. The random variable U is defined by X = 5U.

510152025306 cards, one of each number
(a)

State the distribution of U, including the values it can take.

(1)
(b)

Write down E(U) and find Var(U).

(2)
(c)

Hence find E(X) and Var(X).

(2)
(d)

Find P(X > E(X)).

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
U has a discrete uniform distribution on 1, 2, 3, 4, 5, 6, each with probability 1/6B1
Part (b)
E(U) = (6 + 1)/2 = 3.5B1
Var(U) = (62 − 1)/12 = 35/12B1
Part (c)
E(X) = 5 × 3.5 = 17.5 ftB1
Var(X) = 52 × 35/12 = 875/12 (≈ 72.9) ftB1
Part (d)
X = 20, 25 or 30: 1/2B1

Common error: Writing Var(X) = 5 × 35/12. Multiplying a variable by 5 multiplies its variance by 52 = 25.

9IALM-S1-5-B-05 S1 · ASMedium10 marks

A bag contains 4 red counters and 2 green counters. Two counters are taken from the bag at random, one after the other, without replacement. The random variable X is the number of green counters taken.

(a)

Show that P(X = 1) = 8/15.

(2)
(b)

Find the probability distribution of X.

(2)
(c)

Find E(X).

(1)
(d)

Find Var(X).

(2)
(e)

A player scores 5 points for each green counter taken and loses 2 points for each red counter taken. Write the score S in the form S = aX + b, and find E(S) and Var(S).

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
(4/6)(2/5) + (2/6)(4/5), i.e. both orders countedM1
8/30 + 8/30 = 8/15, as requiredA1
Part (b)
P(X = 0) = (4/6)(3/5) = 2/5B1
P(X = 2) = (2/6)(1/5) = 1/15B1
Part (c)
0 + 8/15 + 2(1/15) = 2/3B1
Part (d)
E(X2) = 8/15 + 4/15 = 4/5, then 4/5 − (2/3)2M1
16/45A1
Part (e)
S = 5X − 2(2 − X) = 7X − 4B1
E(S) = 7(2/3) − 4 = 2/3 ftB1
Var(S) = 49 × 16/45 = 784/45 (≈ 17.4) ftB1

Exam tip: The number of red counters is 2 − X, so the score is a linear function of X. Writing it as aX + b lets you use the rules for E and Var instead of building a new table.

10IALM-S1-5-B-06 S1 · ASMedium8 marks

The number of attempts, X, that a climber needs to complete a route is modelled by P(X = x) = k(5 − x)2 for x = 1, 2, 3, 4, where k is a constant.

(a)

Find the value of k.

(2)
(b)

Find the probability that the climber needs more than one attempt.

(1)
(c)

Find E(X).

(2)
(d)

Find Var(X), giving your answer as an exact fraction.

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
16k + 9k + 4k + k = 1M1
k = 1/30A1
Part (b)
1 − 16/30 = 7/15B1
Part (c)
(1 × 16 + 2 × 9 + 3 × 4 + 4 × 1)/30M1
50/30 = 5/3A1
Part (d)
E(X2) = (16 + 36 + 36 + 16)/30M1
= 52/15A1
Var(X) = 52/15 − 25/9 = 31/45A1

Exam tip: Keep fractions exact all the way through. Rounding E(X) to 1.67 before squaring gives a variance that is slightly wrong.

11IALM-S1-5-B-07 S1 · ASMedium8 marks

In a puzzle app, the level X that a new player reaches on the first day takes the values 1, 2, 3, 4. The cumulative distribution function of X is F(x) = kx(x + 2) for x = 1, 2, 3, 4, where k is a constant.

(a)

Find the value of k.

(2)
(b)

Show that P(X = x) = (2x + 1)/24 for x = 2, 3, 4, and check that this also holds for x = 1.

(3)
(c)

Find E(X) and Var(X).

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
F(4) = 1, so 24k = 1M1
k = 1/24A1
Part (b)
P(X = x) = F(x) − F(x − 1)M1
k[x(x + 2) − (x − 1)(x + 1)] = k(2x + 1) = (2x + 1)/24, as requiredA1
for x = 1: P(X = 1) = P(X ≤ 1) = 3/24, which equals (2 × 1 + 1)/24B1
Part (c)
E(X) = (3 + 10 + 21 + 36)/24 = 35/12B1
E(X2) = (3 + 20 + 63 + 144)/24 = 115/12M1
Var(X) = 115/12 − (35/12)2 = 155/144A1

Exam tip: The first value has no earlier value of F to subtract, so its probability is F at that value; treat it separately when showing a general formula.

12IALM-S1-5-B-08 S1 · ASMedium9 marks

The daily profit, X, in hundreds of pounds, of a market stall has the probability distribution below, where a and b are constants.

x−2013
P(X = x)ab0.30.2
Given that E(X) = 0.5,

(a)

find the value of a and the value of b.

(4)
(b)

Find Var(X).

(3)
(c)

Each day the stallholder records a performance score Y = 4X + 1. Find E(Y) and Var(Y).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
a + b + 0.3 + 0.2 = 1M1
−2a + 0.3 + 0.6 = 0.5M1
a = 0.2A1
b = 0.3A1
Part (b)
E(X2) = 4(0.2) + 1(0.3) + 9(0.2)M1
= 2.9A1
Var(X) = 2.9 − 0.52 = 2.65A1
Part (c)
E(Y) = 4(0.5) + 1 = 3B1
Var(Y) = 16 × 2.65 = 42.4 ftB1

Exam tip: Use the condition Σ P = 1 as one of your equations every time; together with the given mean it fixes both unknowns. Check both answers lie between 0 and 1.

13IALM-S1-5-B-09 S1 · ASMedium9 marks

The number of passengers, X, on a minibus trip has E(X) = 9.2 and Var(X) = 4.36. Each passenger pays a fare of £3.50 and each trip costs £12 to run, so the profit on a trip, in pounds, is R = 3.5X − 12.

(a)

Find E(R).

(2)
(b)

Find the standard deviation of R.

(3)
(c)

Find E(X2).

(2)
(d)

The company changes the fare to £c per passenger, so that the profit becomes cX − 12, with mean £34. Find c and the variance of the new profit.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
3.5 × 9.2 − 12M1
£20.20A1
Part (b)
Var(R) = 3.52 × 4.36M1
= 53.41A1
standard deviation = £7.31 (3 s.f.)A1
Part (c)
E(X2) = Var(X) + [E(X)]2 = 4.36 + 9.22M1
89A1
Part (d)
9.2c − 12 = 34, so c = 5B1
variance = 25 × 4.36 = 109 ftB1

Common error: Subtracting the £12 running cost from the variance. A fixed cost moves the mean only; the spread of the profit comes entirely from 3.5X.

14IALM-S1-5-B-10 S1 · ASMedium9 marks

The 40 tickets in a raffle are numbered 21, 22, 23, …, 60. One ticket is drawn at random, and X is the number on it.

(a)

Write down P(X = x) for x = 21, 22, …, 60.

(1)
(b)

By writing X = U + 20, where U has a discrete uniform distribution on 1, 2, …, 40, find E(X).

(2)
(c)

Find Var(X) and the standard deviation of X.

(3)
(d)

Find the probability that the number drawn is a multiple of 7.

(1)
(e)

Find the probability that X lies within one standard deviation of its mean.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
1/40B1
Part (b)
E(U) = (40 + 1)/2 = 20.5M1
E(X) = 20.5 + 20 = 40.5A1
Part (c)
Var(X) = Var(U) = (402 − 1)/12M1
= 133.25A1
standard deviation = 11.5 (3 s.f.)A1
Part (d)
21, 28, 35, 42, 49, 56: 6/40 = 3/20B1
Part (e)
40.5 ± 11.54…, from 28.96… to 52.04…, so X = 29, …, 52M1
24/40 = 0.6A1

Exam tip: The uniform results (n + 1)/2 and (n2 − 1)/12 are for values 1 to n. Adding 20 to every value adds 20 to the mean and leaves the variance unchanged.

15IALM-S1-5-FB-05 S1 · ASMedium10 marks

A goalkeeper faces two penalty kicks in a practice session. The tree diagram in Figure 1 shows the probabilities of the goalkeeper saving each kick. The random variable X is the number of kicks saved.

1st penalty2nd penalty0.3Saved0.4Saved0.6Not saved0.7Not saved0.2Saved0.8Not saved
(a)

Show that P(X = 1) = 0.32.

(2)
(b)

Find the rest of the probability distribution of X.

(2)
(c)

Find E(X).

(1)
(d)

Find Var(X).

(3)
(e)

The goalkeeper is paid a bonus of £(500X + 100) for the session. Find the mean and the standard deviation of the bonus.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
0.3 × 0.6 + 0.7 × 0.2, i.e. both paths with one saveM1
0.18 + 0.14 = 0.32, as requiredA1
Part (b)
P(X = 0) = 0.7 × 0.8 = 0.56B1
P(X = 2) = 0.3 × 0.4 = 0.12B1
Part (c)
0.32 + 2(0.12) = 0.56B1
Part (d)
E(X2) = 0.32 + 4(0.12)M1
= 0.8A1
Var(X) = 0.8 − 0.562 = 0.4864A1
Part (e)
mean = 500 × 0.56 + 100 = £380 ftB1
s.d. = 500 × √0.4864 = £349 (3 s.f.) ftB1

Exam tip: The standard deviation of aX + b is |a| × the standard deviation of X: the added constant does not spread the values out at all.

16IALM-S1-5-FB-06 S1 · ASMedium9 marks

The number of people, X, in a car arriving at a car park is modelled by the probability distribution shown in Figure 1, where k is a constant. The car park charges £4 per car plus £1.50 per person, so the charge for a car, in pounds, is C = 1.5X + 4.

12345xP(X = x)0k2k3k4k5kk3k4k3kk
(a)

Find the value of k.

(2)
(b)

Explain, without calculation, why E(X) = 3.

(1)
(c)

Find Var(X).

(3)
(d)

Find E(C) and Var(C).

(2)
(e)

Find the probability that the charge for a car is more than £8.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
k + 3k + 4k + 3k + k = 1M1
k = 1/12A1
Part (b)
the distribution is symmetrical about x = 3, so the mean is 3B1
Part (c)
E(X2) = (1 + 4 × 3 + 9 × 4 + 16 × 3 + 25)/12M1
= 122/12 = 61/6A1
Var(X) = 61/6 − 9 = 7/6A1
Part (d)
E(C) = 1.5 × 3 + 4 = £8.50B1
Var(C) = 1.52 × 7/6 = 2.625 ftB1
Part (e)
1.5X + 4 > 8 gives X > 8/3, so X = 3, 4, 5: 8/12 = 2/3 ftB1

Exam tip: When the bars are symmetrical about one value, that value is the mean. Saying so saves a calculation, but you still need E(X2) for the variance.

17IALM-S1-5-FB-07 S1 · ASMedium10 marks

Figure 1 shows a circular target at a fair, made from three concentric circles of radii 10 cm, 20 cm and 30 cm. A dart thrown by a beginner always lands on the target and is equally likely to land at any point of it, so the probability of landing in a region is proportional to its area. The score, X, is 10, 5 or 2 as shown.

105210 cm20 cm30 cmRadii 10 cm, 20 cmand 30 cm (to scale)Scores:inner disc 10middle ring 5outer ring 2
(a)

Show that P(X = 10) = 1/9.

(2)
(b)

Find P(X = 5) and P(X = 2).

(1)
(c)

Find E(X).

(2)
(d)

Find Var(X).

(3)
(e)

A player pays 40p for one dart and receives 10p for each point scored. Find the expected profit, in pence, made by the stall on each dart thrown.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
area of inner disc ÷ area of target = π(10)2 ÷ π(30)2M1
= 100/900 = 1/9, as requiredA1
Part (b)
(400 − 100)/900 = 1/3 and (900 − 400)/900 = 5/9B1
Part (c)
10(1/9) + 5(3/9) + 2(5/9)M1
= 35/9 (≈ 3.89)A1
Part (d)
E(X2) = 100/9 + 25(3/9) + 4(5/9)M1
= 195/9 = 65/3A1
Var(X) = 65/3 − (35/9)2 = 530/81 (≈ 6.54)A1
Part (e)
profit = 40 − 10X, so expected profit = 40 − 10 × 35/9M1
= 10/9 ≈ 1.11p ftA1

Common error: Taking the probabilities as 1/3 each because the rings are equally wide. Probability goes with area, and area grows with the square of the radius.

18IALM-S1-5-FB-08 S1 · ASMedium9 marks

Spinner A has three equal sectors numbered 1, 2, 3 and spinner B has four equal sectors numbered 1, 2, 3, 4. Both are fair and are spun once, independently. The random variable X is the positive difference between the two scores (0 if they are equal). Figure 1 shows the value of X for each of the 12 equally likely outcomes.

Spinner BSpinner A1234101232101232101Each cell shows the value of X
(a)

Find the probability distribution of X.

(3)
(b)

Find E(X).

(2)
(c)

Find Var(X).

(3)
(d)

Find the probability that X is greater than its mean.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
each cell has probability 1/12, and the cells for each value are countedM1
P(X = 1) = 5/12A1
P(X = 0) = 1/4, P(X = 2) = 1/4, P(X = 3) = 1/12A1
Part (b)
(0 × 3 + 1 × 5 + 2 × 3 + 3 × 1)/12M1
= 14/12 = 7/6A1
Part (c)
E(X2) = (5 + 4 × 3 + 9 × 1)/12M1
= 26/12 = 13/6A1
Var(X) = 13/6 − (7/6)2 = 29/36A1
Part (d)
X > 7/6 means X = 2 or 3: 4/12 = 1/3 ftB1

Exam tip: A sample-space grid only gives probabilities by counting when every cell is equally likely. Here both spinners are fair and independent, so each of the 12 cells has probability 1/12.

19IALM-S1-5-FB-09 S1 · ASMedium9 marks

The number of faults, X, found in a weekly inspection of a machine takes the values 0, 1, 2 and 3. Figure 1 shows the cumulative distribution function F(x) of X, where a and b are constants. The graph is not drawn to scale: the levels a and b are labelled on the vertical axis, and their values are to be found from the information below, not read from the graph. It is known that E(X) = 1.9 and that P(X = 3) = 2P(X = 1).

01234xF(x)0.1ab10Not drawn to scale
(a)

Write down expressions in terms of a and b for P(X = 1), P(X = 2) and P(X = 3).

(2)
(b)

Show that a + b = 1.

(2)
(c)

Find the value of a and the value of b.

(2)
(d)

Find Var(X).

(3)
Show mark scheme
AnswerMarkNotes
Part (a)
P(X = 1) = a − 0.1 and P(X = 2) = b − aB1
P(X = 3) = 1 − bB1
Part (b)
(a − 0.1) + 2(b − a) + 3(1 − b) = 1.9M1
2.9 − a − b = 1.9, so a + b = 1, as requiredA1
Part (c)
1 − b = 2(a − 0.1), i.e. 2a + b = 1.2, solved with a + b = 1M1
a = 0.2, b = 0.8A1
Part (d)
probabilities 0.1, 0.1, 0.6, 0.2, so E(X2) = 0.1 + 4(0.6) + 9(0.2)M1
= 4.3A1
Var(X) = 4.3 − 1.92 = 0.69A1

Exam tip: Turn every fact into an equation in the unknowns: here the graph gives the probabilities as differences of F, and the mean and the ratio give two equations for a and b. Check the answers keep F increasing.

20IALM-S1-5-FB-10 S1 · ASMedium9 marks

A game uses two six-sided dice, each numbered 1 to 6. The score X on the fair die and the score Y on a biased die have the distributions shown in Figure 1. The probabilities for the biased die lie on gridlines.

00.050.100.150.200.250.300.35123456Fair die: score Xscoreprobabilityeach height 1/600.050.100.150.200.250.300.35123456Biased die: score Yscoreprobability
(a)

Name the distribution of X, and find E(X) and Var(X).

(3)
(b)

Find E(Y).

(2)
(c)

Find Var(Y).

(3)
(d)

State, with a reason, which die gives the more variable score.

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
discrete uniform distribution on 1, 2, …, 6B1
E(X) = (6 + 1)/2 = 3.5B1
Var(X) = (62 − 1)/12 = 35/12 (≈ 2.92)B1
Part (b)
1(0.1) + 2(0.1) + 3(0.15) + 4(0.15) + 5(0.2) + 6(0.3)M1
4.15A1
Part (c)
E(Y2) = 0.1 + 0.4 + 1.35 + 2.4 + 5 + 10.8M1
= 20.05A1
Var(Y) = 20.05 − 4.152 = 2.8275 (≈ 2.83)A1
Part (d)
the fair die, since Var(X) = 2.92 > Var(Y) = 2.83 ftB1

Exam tip: For a discrete uniform distribution on 1, 2, …, n you can quote E(X) = (n + 1)/2 and Var(X) = (n2 − 1)/12 instead of building a table.

21IALM-S1-5-B-11 S1 · ASHard12 marks

In a game, a fair four-sided die numbered 1, 2, 3, 4 is rolled until a 4 is obtained, with a maximum of three rolls: the player stops after the third roll whatever it shows. The random variable X is the number of rolls made.

(a)

Show that P(X = 3) = 9/16.

(2)
(b)

Find the probability distribution of X, and find P(X ≤ 2).

(3)
(c)

Find E(X).

(2)
(d)

Find Var(X).

(3)
(e)

Each roll costs 20 pence. Find the mean and the standard deviation of the total cost, in pence, of one game.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
X = 3 exactly when the first two rolls are not 4M1
(3/4)2 = 9/16, as requiredA1
Part (b)
P(X = 1) = 1/4B1
P(X = 2) = (3/4)(1/4) = 3/16B1
P(X ≤ 2) = 1/4 + 3/16 = 7/16 ftB1
Part (c)
1(1/4) + 2(3/16) + 3(9/16)M1
37/16A1
Part (d)
E(X2) = 1(1/4) + 4(3/16) + 9(9/16)M1
= 97/16A1
Var(X) = 97/16 − (37/16)2 = 183/256 (≈ 0.715)A1
Part (e)
mean = 20 × 37/16 = 46.25 p ftB1
standard deviation = 20√(183/256) = 16.9 p (3 s.f.) ftB1

Exam tip: Read the stopping rule carefully: X = 3 includes every game in which the third roll is reached, whether or not it shows a 4.

22IALM-S1-5-B-12 S1 · ASHard11 marks

The number of loaves, X, bought by a customer at a bakery is modelled by P(X = x) = a + bx for x = 1, 2, 3, 4, where a and b are constants. The mean number of loaves bought is 2.75.

(a)

Show that 4a + 10b = 1 and 10a + 30b = 2.75.

(3)
(b)

Find the value of a and the value of b.

(2)
(c)

Find Var(X).

(3)
(d)

Find P(X < 2.5).

(1)
(e)

The bakery's points scheme gives a customer who buys x loaves 2x points. Find the expected number of points per customer.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
Σ(a + bx) = 4a + (1 + 2 + 3 + 4)b = 1, giving 4a + 10b = 1B1
E(X) = Σ x(a + bx)M1
(1 + 2 + 3 + 4)a + (1 + 4 + 9 + 16)b = 10a + 30b = 2.75, as requiredA1
Part (b)
eliminates one unknown, e.g. 2.5 × first: 10a + 25b = 2.5, so 5b = 0.25M1
b = 0.05, a = 0.125A1
Part (c)
probabilities 0.175, 0.225, 0.275, 0.325, and E(X2) = Σ x2P(X = x)M1
E(X2) = 8.75A1
Var(X) = 8.75 − 2.752 = 1.1875A1
Part (d)
P(X ≤ 2) = 0.175 + 0.225 = 0.4B1
Part (e)
E(2X) = 2(0.175) + 4(0.225) + 8(0.275) + 16(0.325)M1
8.65A1

Exam tip: E(2X) means apply 2x to each value and then average: it is not 22.75.

23IALM-S1-5-B-13 S1 · ASHard12 marks

A fair spinner has n equal sectors numbered 1, 2, …, n. The score X on one spin has a discrete uniform distribution. It is known that E(X2) = 63.

(a)

Using Σr2 = n(n + 1)(2n + 1)/6, show that E(X2) = (n + 1)(2n + 1)/6, and hence find n.

(4)
(b)

Find E(X) and Var(X).

(2)
(c)

In a game, a player's score is S = 5X − 20. Find E(S) and Var(S).

(2)
(d)

Find P(S > 0).

(2)
(e)

The spinner is spun twice. Find the probability that exactly one of the two scores S is positive.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
E(X2) = (1/n)Σr2 = (n + 1)(2n + 1)/6B1
(n + 1)(2n + 1) = 378, so 2n2 + 3n − 377 = 0M1
(2n + 29)(n − 13) = 0, or the quadratic formulaM1
n = 13, since n > 0A1
Part (b)
E(X) = 7B1
Var(X) = 63 − 49 = 14B1
Part (c)
E(S) = 5 × 7 − 20 = 15 ftB1
Var(S) = 25 × 14 = 350 ftB1
Part (d)
5X − 20 > 0 gives X > 4, so X = 5, …, 13M1
9/13A1
Part (e)
2 × (9/13) × (4/13)M1
72/169 ftA1

Exam tip: Once n is known, the variance follows from E(X2) − [E(X)]2 or from (n2 − 1)/12; both give 14, which is a useful check.

24IALM-S1-5-B-14 S1 · ASHard14 marks

A student answers three quiz questions, A, B and C. The probabilities of answering them correctly are 0.9, 0.7 and 0.5 respectively, independently of each other. The random variable X is the number of questions answered correctly.

A0.9B0.7C0.5C′0.5B′0.3C0.5C′0.5A′0.1B0.7C0.5C′0.5B′0.3C0.5C′0.5
Figure: tree diagram for the three questions (A = question A answered correctly, A′ = not)
(a)

Show that P(X = 2) = 0.485.

(3)
(b)

Find the rest of the probability distribution of X, and hence find P(X ≤ 1).

(3)
(c)

Find E(X) and Var(X).

(4)
(d)

The student gains 4 points for each correct answer and loses 1 point for each incorrect answer. Show that the total score is T = 5X − 3, and find E(T) and Var(T).

(3)
(e)

Find the probability that the student's score is greater than E(T).

(1)
Show mark scheme
AnswerMarkNotes
Part (a)
identifies the three cases: A and B only, A and C only, B and C onlyM1
0.9 × 0.7 × 0.5 + 0.9 × 0.3 × 0.5 + 0.1 × 0.7 × 0.5M1
0.315 + 0.135 + 0.035 = 0.485, as requiredA1
Part (b)
P(X = 0) = 0.1 × 0.3 × 0.5 = 0.015 and P(X = 3) = 0.9 × 0.7 × 0.5 = 0.315B1
P(X = 1) = 1 − (0.015 + 0.485 + 0.315) = 0.185B1
P(X ≤ 1) = 0.015 + 0.185 = 0.2 ftB1
Part (c)
0.185 + 2(0.485) + 3(0.315)M1
E(X) = 2.1A1
E(X2) = 0.185 + 4(0.485) + 9(0.315) = 4.96M1
Var(X) = 4.96 − 2.12 = 0.55A1
Part (d)
T = 4X − (3 − X) = 5X − 3, as requiredB1
E(T) = 5 × 2.1 − 3 = 7.5 ftB1
Var(T) = 25 × 0.55 = 13.75 ftB1
Part (e)
5X − 3 > 7.5 gives X > 2.1, so X = 3: 0.315B1

Exam tip: When a list of probabilities must add to 1, find the hardest one by subtraction, but only after the others are certain; then check it directly if time allows.

25IALM-S1-5-FB-11 S1 · ASHard12 marks

Figure 1 shows the six balls in a tub at a school fair. A player takes two balls at random from the tub, without replacement. The random variable X is the sum of the numbers on the two balls.

1112236 balls in a tub
(a)

Show that P(X = 4) = 4/15.

(3)
(b)

Find the rest of the probability distribution of X.

(2)
(c)

Find E(X).

(2)
(d)

Find Var(X).

(3)
(e)

The player is awarded W = 3X − 4 points. Find E(W) and the standard deviation of W.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
a 1 and the 3, in either order: 2 × (3/6)(1/5) = 1/5M1
two 2s: (2/6)(1/5) = 1/15M1
1/5 + 1/15 = 4/15, as requiredA1
Part (b)
P(X = 2) = (3/6)(2/5) = 1/5B1
P(X = 3) = 2(3/6)(2/5) = 2/5 and P(X = 5) = 2(2/6)(1/5) = 2/15B1
Part (c)
2(1/5) + 3(2/5) + 4(4/15) + 5(2/15)M1
= 50/15 = 10/3A1
Part (d)
E(X2) = 4(3/15) + 9(6/15) + 16(4/15) + 25(2/15)M1
= 180/15 = 12A1
Var(X) = 12 − (10/3)2 = 8/9A1
Part (e)
E(W) = 3 × 10/3 − 4 = 6 ftB1
s.d. = 3 × √(8/9) = 2√2 ≈ 2.83 ftB1

Common error: Counting a 1 and a 3 only once. Taking the 1 first and taking the 3 first are different outcomes, so that probability is doubled; two balls of the same number appear only once.

26IALM-S1-5-FB-12 S1 · ASHard12 marks

Figure 1 shows a game at a fair. A ball is released at the top of the board and meets one pin in each of the four rows. At each pin it is equally likely to go to the left or to the right, independently of what happened at earlier pins, and it then falls into one of the slots A to E. The number of points scored, X, is shown on each slot.

ball released hereA10 ptsB2 ptsC0 ptsD2 ptsE10 pts
(a)

Show that the probability that the ball lands in slot B is 1/4.

(2)
(b)

Find the probability that the ball lands in slot C.

(2)
(c)

Find the probability distribution of X.

(2)
(d)

Find E(X) and Var(X).

(4)
(e)

The stall pays out 20p for each point scored. Find the charge per go, in pence, so that the stall's expected profit per go is 15p.

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
each route through the four pins has probability (1/2)4 = 1/16M1
slot B needs exactly one right turn: 4 routes, so 4/16 = 1/4, as requiredA1
Part (b)
routes with exactly two right turns: RRLL, RLRL, RLLR, LRRL, LRLR, LLRRM1
6/16 = 3/8A1
Part (c)
P(X = 10) = 2/16 = 1/8 (slots A and E)B1
P(X = 2) = 8/16 = 1/2 and P(X = 0) = 3/8 ftB1
Part (d)
E(X) = 10(1/8) + 2(1/2) = 2.25B1
E(X2) = 100(1/8) + 4(1/2)M1
= 14.5A1
Var(X) = 14.5 − 2.252 = 9.4375 (≈ 9.44)A1
Part (e)
c − 20 × 2.25 = 15M1
c = 60p ftA1

Exam tip: On a pin board only the number of right turns decides the slot, not their order. List the routes systematically (or use the pattern 1, 4, 6, 4, 1) and check the counts add up to 24 = 16.

27IALM-S1-5-FB-13 S1 · ASHard11 marks

The number of parcels, X, delivered to an office in a randomly chosen hour takes the values 0, 1, 2, 3 and 4. Figure 1 shows part of the probability distribution of X: the probabilities shown lie on gridlines, and P(X = 1) = p and P(X = 3) = q are not drawn. It is known that E(X2) = 5.65.

00.050.10.150.20.250.30.3501234xP(X = x)pqlines for x = 1 and x = 3 not shown
(a)

Show that p + q = 0.45.

(1)
(b)

Find the value of p and the value of q.

(3)
(c)

Find E(X) and Var(X).

(3)
(d)

The office manager uses the score Y = 20 − 4X. Find E(Y) and Var(Y).

(2)
(e)

Find P(Y > E(Y)).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
0.1 + p + 0.3 + q + 0.15 = 1, so p + q = 0.45, as requiredB1
Part (b)
E(X2): p + 4(0.3) + 9q + 16(0.15) = 5.65, so p + 9q = 2.05M1
subtracting p + q = 0.45: 8q = 1.6M1
q = 0.2, p = 0.25A1
Part (c)
E(X) = 0.25 + 2(0.3) + 3(0.2) + 4(0.15) = 2.05 ftB1
Var(X) = 5.65 − 2.052M1
= 1.4475 (≈ 1.45)A1
Part (d)
E(Y) = 20 − 4 × 2.05 = 11.8 ftB1
Var(Y) = (−4)2 × 1.4475 = 23.16 (≈ 23.2) ftB1
Part (e)
20 − 4X > 11.8 gives X < 2.05, i.e. X ≤ 2M1
0.1 + 0.25 + 0.3 = 0.65 ftA1

Common error: Writing Var(Y) = −4 × 1.4475 (a negative variance) or 4 × 1.4475. The constant multiplying X is squared: (−4)2 = 16.

28IALM-S1-5-FB-14 S1 · ASHard12 marks

In a computer game, a player opening a chest receives X coins. Figure 1 shows the cumulative distribution function F(x) of X. Every step of the graph lies on a gridline.

00.10.20.30.40.50.60.70.80.91.00246810121416182022xF(x)
(a)

Write down F(9) and find P(6 ≤ X ≤ 12).

(2)
(b)

Explain why X = 2U, where U has a discrete uniform distribution on 1, 2, …, 10, and write down P(X = x).

(2)
(c)

Find E(X) and Var(X).

(3)
(d)

The game converts coins into a score Y = aX + b, where a and b are constants and a > 0. Given that E(Y) = 50 and Var(Y) = 297, find a and b.

(3)
(e)

Find P(Y > 60).

(2)
Show mark scheme
AnswerMarkNotes
Part (a)
F(9) = F(8) = 0.4B1
F(12) − F(4) = 0.6 − 0.2 = 0.4B1
Part (b)
F jumps only at x = 2, 4, …, 20 and every jump is 0.1, so the 10 values 2u are equally likelyB1
P(X = x) = 0.1 for x = 2, 4, 6, …, 20B1
Part (c)
E(U) = 11/2, so E(X) = 2 × 5.5 = 11B1
Var(U) = (102 − 1)/12 = 8.25M1
Var(X) = 22 × 8.25 = 33A1
Part (d)
a2 × 33 = 297M1
a2 = 9 and a > 0, so a = 3A1
3 × 11 + b = 50, so b = 17A1
Part (e)
3X + 17 > 60 gives X > 43/3 ≈ 14.3, so X = 16, 18 or 20M1
0.3 ftA1

Exam tip: F keeps the same value between the points where it jumps, so F(9) is just F(8). For whole-number questions on a step graph, find the last jump at or before the value you need.