A Level · Mathematics 9709 · P5

A Level Mathematics Paper 5 · Probability & Statistics 1 practice papers

50 marks · 1h 15m. Each generated paper follows the 9709 Paper 5 · Probability & Statistics 1 format, with the same number and style of questions as the real exam, and comes with a Cambridge-style mark scheme.

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Sample Paper 5 · Probability & Statistics 1 questions

Question 1Structured · 7 marksA* challenge

Bag X contains 3 red and 5 blue discs. Bag Y contains 6 red and 3 blue discs. A biased coin, with P(head) = 0.6, is thrown. If it shows a head, a disc is taken at random from bag X; otherwise a disc is taken at random from bag Y.

(a)

Find the probability that the disc is red.

[2]
(b)

Given that the disc is red, find the probability that it was taken from bag X.

[3]
(c)

Determine whether the events "the coin shows a head" and "the disc is red" are independent. Justify your answer.

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)0.4917
sum of two products M1
0.4917 A1
2
(b)0.4576
P(X and red) ÷ P(red) M1
0.225 ÷ 0.4917 A1
0.4576 A1
3
(c)not independent: P(red | head) = 0.375 ≠ P(red) = 0.4917
valid comparison of probabilities M1
correct conclusion A1
2
Question 2Structured · 6 marksA* challenge

A bag contains 6 discs numbered 1 to 6. Two discs are taken at random, without replacement. The random variable X is the larger of the two numbers.

(a)

Draw up the probability distribution table for X.

[3]
(b)

Find E(X) and Var(X).

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)P(X = 2) = 115, P(X = 3) = 215, P(X = 4) = 15, P(X = 5) = 415, P(X = 6) = 13
correct values of X B1
at least two correct probabilities M1
all correct A1
3
(b)E(X) = 143, Var(X) = 1.556
Σxp M1
Σx2p − (E(X))2 M1
both correct A1
3
Question 3Structured · 4 marks

Events A and B satisfy P(A) = 0.5, P(B) = 0.25 and P(A ∩ B) = 0.125.

(a)

Show that A and B are independent.

[1]
(b)

Find P(A ∪ B) and P(A | B′).

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)P(A)P(B) = P(A ∩ B)
0.5 × 0.25 = 0.125 B1
1
(b)0.625; 0.5
P(A ∪ B) = 0.625 B1
conditional formula M1
0.5 A1
3
Question 4Structured · 5 marks

The masses, X g, of apples are normally distributed with mean 52 and standard deviation 4.

(a)

Find P(X < 52.8).

[3]
(b)

Find P(52.8 < X < 54.4).

[2]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)0.5793
Standardise M1
z = 0.200 A1
0.5793 A1
3
(b)0.1465
Φ(z2) − Φ(z1) M1
0.1465 A1
2
Question 5Structured · 3 marks

For a set of 27 values, Σx = 1158.3 and Σx2 = 51376.1. Find the mean and standard deviation.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
mean 42.9, sd 7.9
42.9 B1
√(Σx2/n − mean2) M1
7.9 A1
3
Question 6Structured · 5 marks

The random variable X has the probability distribution shown.

x1234
P(X = x)0.120.170.18k
Table 6.1
(a)

Find k.

[1]
(b)

Find E(X) and Var(X).

[4]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)0.53
0.53 B1
1
(b)E(X) = 3.12, Var(X) = 1.17
Σxp M1
3.12 A1
Σx2p − E(X)2 M1
1.17 A1
4

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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in