Statistics: Probability - Worksheets, Questions and Revision

12 original exam-style questions - 3 pages of questions with a full mark scheme - free printable PDF.

Download PDF Jump to mark scheme (page 4)Read the revision guide

Next in this strand

Revision Library

revisionlibrary.co.uk

A-Level · Statistics

S3 Statistics: Probability

EDEXCEL 9MA0 · Calculator allowed · about 145 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A machine in a factory produces items in batches of 20. A quality inspector models the number of defective items in a randomly chosen batch, D, using the distribution D ~ B(20, 0.05).

(a)State two assumptions needed for this binomial model to be a suitable model for D.(2)

(b)Using this model, calculate P(D = 2), giving your answer to 3 significant figures.(3)

(Total for Question 1 is 5 marks)

At a sixth form college, 120 students were surveyed about the subjects they study. Let M be the event that a randomly chosen student studies Mathematics and P be the event that a randomly chosen student studies Physics. From the survey, P(M) = 0.65, P(P) = 0.40 and P(M and P) = 0.25.

(a)Find P(M or P).(2)

(b)Find P(M and not P), the probability that a randomly chosen student studies Mathematics but not Physics.(2)

(c)Determine, showing your working, whether the events M and P are independent.(3)

(d)Find P(not M and not P).(2)

(Total for Question 2 is 9 marks)

For two events A and B, P(A) = 0.3, P(B) = 0.45 and P(A or B) = 0.65.

(a)Find P(A and B).(2)

(b)Show that A and B are not independent.(3)

(c)State, with a reason, whether A and B could be mutually exclusive.(2)

(Total for Question 3 is 7 marks)

A bag contains 5 red counters and 7 blue counters. Two counters are drawn at random, one after another, without replacement.

(a)Find the probability that both counters drawn are the same colour.(3)

(b)Find the probability that at least one of the two counters drawn is red.(3)

(c)Given that both counters drawn are the same colour, find the probability that both are red.(3)

(Total for Question 4 is 9 marks)

The random variable X ~ B(10, 0.3).

(a)Calculate P(X = 3), giving your answer to 3 significant figures.(2)

(b)Calculate P(X ≤ 2).(3)

(c)Calculate P(X ≥ 4).(3)

(Total for Question 5 is 8 marks)

For events A and B, P(A) = 0.4, P(B) = 0.35 and P(A | B) = 0.6.

(a)Find P(A and B).(2)

(b)Find P(B | A).(2)

(c)Find P(A or B).(2)

(d)Determine whether A and B are independent.(2)

(Total for Question 6 is 8 marks)

The discrete random variable X has probability distribution given by P(X = x) = kx for x = 1, 2, 3, 4, and P(X = x) = 0 otherwise, where k is a constant.

(a)Show that k = 0.1.(2)

(b)Find P(X ≥ 3).(2)

(c)Find E(X).(3)

(d)Find Var(X).(4)

(Total for Question 7 is 11 marks)

In a survey of 200 people, C is the event that a randomly chosen person owns a car and Bi is the event that a randomly chosen person owns a bicycle. It is known that P(C) = 0.55, P(Bi) = 0.30 and P(neither a car nor a bicycle) = 0.20.

(a)Show that P(C and Bi) = 0.05.(3)

(b)Find the number of the 200 people who own a bicycle but not a car.(2)

(c)Two people are selected at random, without replacement, from those in the survey who own a car. Find the probability that both of the people selected also own a bicycle.(4)

(Total for Question 8 is 9 marks)

A factory uses two machines to produce components. Machine A produces 60% of all components and Machine B produces the remaining 40%. Machine A produces defective components at a rate of 3%, and Machine B produces defective components at a rate of 5%.

(a)Find the probability that a randomly selected component is defective.(3)

(b)Given that a component is defective, find the probability that it was produced by Machine B.(3)

(c)Two components are selected independently at random. Find the probability that exactly one of the two components is defective.(3)

(Total for Question 9 is 9 marks)

The random variable X ~ B(n, p). Given that E(X) = 6 and Var(X) = 4.2.

(a)Show that p = 0.3.(3)

(b)Find the value of n.(2)

(c)Find P(X = 8).(3)

(Total for Question 10 is 8 marks)

Events A and B are independent, with P(A) = p and P(B) = p + 0.2, where 0 < p < 0.8. Given that P(A or B) = 0.75.

(a)Show that p satisfies the equation p² - 1.8p + 0.55 = 0.(4)

(b)Hence find the value of p, giving a reason why the other root of the equation should be rejected.(3)

(c)State, with a reason, the value of P(A | B').(2)

(Total for Question 11 is 9 marks)

A box contains n balls, of which 3 are red and the rest are blue, where n > 3. Two balls are drawn at random from the box, without replacement. Given that P(both balls drawn are red) = 1/15.

(a)Show that n² - n - 90 = 0.(4)

(b)Hence show that n = 10, explaining why the other solution of the equation is rejected.(2)

(c)Given that n = 10, find the probability that exactly one of the two balls drawn is red.(3)

(d)A third ball is now drawn at random from the box, without replacement, given that the first two balls drawn were both red. Find the probability that the third ball is blue.(3)

(Total for Question 12 is 12 marks)

Mark scheme · S3 Statistics: Probability

Question 1

(a) B1 the probability that an item is defective is constant for every item in the batch, oe
(a) B1 whether one item is defective is independent of whether any other item is defective, oe
(a) Answer: Any two of: constant probability of an item being defective for each item; items are defective independently of one another; fixed number of items n = 20; only two outcomes (defective or not defective)
(b) M1 use of binomial formula with correct combination, e.g. 20C2 x 0.05² x 0.95¹⁸
(b) M1 correct substitution: 190 x 0.0025 x 0.95¹⁸
(b) A1 awrt 0.189
(b) Answer: P(D = 2) = 0.189 (3 s.f.)

Question 2

(a) M1 P(M) + P(P) - P(M and P)
(a) A1 cao 0.80
(a) Answer: P(M or P) = 0.80
(b) M1 P(M) - P(M and P)
(b) A1 cao 0.40
(b) Answer: P(M and not P) = 0.40
(c) M1 compute P(M) x P(P) = 0.65 x 0.40
(c) A1 P(M) x P(P) = 0.26
(c) A1 cso: 0.26 does not equal 0.25 (=P(M and P)), so M and P are not independent
(c) Answer: Not independent, since P(M)P(P) = 0.26 not equal to P(M and P) = 0.25
(d) M1ft 1 - P(M or P), following through from part (a)
(d) A1ft cao 0.20
(d) Answer: P(not M and not P) = 0.20

Question 3

(a) M1 P(A) + P(B) - P(A or B)
(a) A1 cao 0.10
(a) Answer: P(A and B) = 0.10
(b) M1 compute P(A) x P(B) = 0.3 x 0.45
(b) A1 P(A) x P(B) = 0.135, compared with P(A and B) = 0.10 (ft from a)
(b) A1 cso: since 0.135 does not equal 0.10, A and B are not independent
(b) Answer: Not independent, since P(A)P(B) = 0.135 not equal to P(A and B) = 0.10
(c) B1 No
(c) B1 reason: since P(A and B) = 0.10, which is not equal to 0 (mutually exclusive events would require P(A and B) = 0), oe
(c) Answer: No, since P(A and B) = 0.10 not equal to 0

Question 4

(a) M1 P(RR) = 5/12 x 4/11, oe
(a) M1 P(BB) = 7/12 x 6/11, oe
(a) A1 cao 31/66 (awrt 0.470)
(a) Answer: P(same colour) = 31/66 (awrt 0.470)
(b) M1 P(BB) = 7/12 x 6/11
(b) M1 1 - P(BB)
(b) A1 cao 15/22 (awrt 0.682)
(b) Answer: P(at least one red) = 15/22 (awrt 0.682)
(c) M1 identify P(RR) = 5/33 (ft from a)
(c) M1 divide by P(same colour) = 31/66 (ft from a)
(c) A1 cao 10/31 (awrt 0.323)
(c) Answer: P(both red | same colour) = 10/31 (awrt 0.323)

Question 5

(a) M1 10C3 x 0.3³ x 0.7⁷
(a) A1 awrt 0.267
(a) Answer: P(X = 3) = 0.267 (3 s.f.)
(b) M1 P(X=0) + P(X=1) + P(X=2)
(b) M1 correct individual terms: 0.7¹⁰ + 10x0.3x0.7⁹ + 45x0.3^2x0.7^8
(b) A1 awrt 0.383
(b) Answer: P(X ≤ 2) = 0.383 (3 s.f.)
(c) M1 1 - P(X ≤ 3)
(c) M1 P(X ≤ 3) = P(X ≤ 2) (ft from b) + P(X = 3) (ft from a)
(c) A1 awrt 0.350
(c) Answer: P(X ≥ 4) = 0.350 (3 s.f.)

Question 6

(a) M1 P(A|B) x P(B)
(a) A1 cao 0.21
(a) Answer: P(A and B) = 0.21
(b) M1ft P(A and B) / P(A), following through from part (a)
(b) A1 cao 0.525
(b) Answer: P(B | A) = 0.525
(c) M1ft P(A) + P(B) - P(A and B), following through from part (a)
(c) A1 cao 0.54
(c) Answer: P(A or B) = 0.54
(d) M1 compute P(A) x P(B) = 0.4 x 0.35 = 0.14
(d) A1 cso: since 0.14 does not equal 0.21 (=P(A and B) from a), A and B are not independent
(d) Answer: Not independent, since P(A)P(B) = 0.14 not equal to P(A and B) = 0.21

Question 7

(a) M1 sum of probabilities = 1: k(1+2+3+4) = 1, i.e. 10k = 1
(a) A1 cso: k = 0.1
(a) Answer: k = 0.1
(b) M1 P(X=3) + P(X=4) = 3k + 4k, using k = 0.1 (ft)
(b) A1 cao 0.7
(b) Answer: P(X ≥ 3) = 0.7
(c) M1 sum of x P(x) with at least three correct terms
(c) M1 all four terms correct: 1(0.1)+2(0.2)+3(0.3)+4(0.4)
(c) A1 cao 3.0
(c) Answer: E(X) = 3.0
(d) M1 attempt E(X²) = sum x² P(x)
(d) M1 correct value E(X²) = 10.0
(d) M1ft Var(X) = E(X²) - [E(X)]², following through from c
(d) A1 cao 1.0
(d) Answer: Var(X) = 1.0

Question 8

(a) M1 P(C or Bi) = 1 - P(neither) = 0.80
(a) M1 use P(C or Bi) = P(C) + P(Bi) - P(C and Bi), rearranged for P(C and Bi)
(a) A1 cso: 0.55 + 0.30 - 0.80 = 0.05
(a) Answer: P(C and Bi) = 0.05
(b) M1 P(Bi and not C) = P(Bi) - P(C and Bi) = 0.30 - 0.05, then x 200 (ft from a)
(b) A1 cao 50
(b) Answer: 50 people
(c) M1 number owning a car = 0.55 x 200 = 110
(c) M1 number owning both a car and a bicycle = 0.05 x 200 = 10 (ft from a)
(c) M1 (10/110) x (9/109), oe
(c) A1 cao 9/1199 (awrt 0.00751)
(c) Answer: P(both own a bicycle) = 9/1199 (awrt 0.00751)

Question 9

(a) M1 P(D) = P(A)P(D|A) + P(B)P(D|B)
(a) M1 correct substitution: 0.6 x 0.03 + 0.4 x 0.05
(a) A1 cao 0.038
(a) Answer: P(defective) = 0.038
(b) M1 P(B and D) = 0.4 x 0.05 = 0.02
(b) M1ft P(B and D) / P(D), following through from part (a)
(b) A1 cao 10/19 (awrt 0.526)
(b) Answer: P(Machine B | defective) = 10/19 (awrt 0.526)
(c) M1 2 x P(D) x P(D'), oe, using P(D) ft from a
(c) M1 correct substitution: 2 x 0.038 x 0.962
(c) A1 awrt 0.0731
(c) Answer: P(exactly one defective) = 0.0731 (3 s.f.)

Question 10

(a) M1 np = 6 and np(1-p) = 4.2
(a) M1 divide to eliminate n: (1-p) = 4.2/6 = 0.7
(a) A1 cso p = 0.3
(a) Answer: p = 0.3
(b) M1ft n = 6 / 0.3, following through from part (a)
(b) A1 cao n = 20
(b) Answer: n = 20
(c) M1ft 20C8 x 0.3⁸ x 0.7¹², using n and p ft from a, b
(c) M1 correct substitution of values
(c) A1 awrt 0.114
(c) Answer: P(X = 8) = 0.114 (3 s.f.)

Question 11

(a) M1 use P(A or B) = P(A) + P(B) - P(A)P(B), using independence
(a) M1 correct substitution: p + (p+0.2) - p(p+0.2) = 0.75
(a) dM1 expand and rearrange into a three-term quadratic set to zero
(a) A1 cso: p² - 1.8p + 0.55 = 0, fully shown
(a) Answer: p² - 1.8p + 0.55 = 0 (shown)
(b) M1 solve p² - 1.8p + 0.55 = 0 using the quadratic formula or a calculator
(b) A1 awrt p = 0.390
(b) B1 reason: the other root (awrt 1.41) is rejected as it exceeds 1 and probabilities cannot be greater than 1, oe
(b) Answer: p = 0.390 (3 s.f.); other root p = 1.41 rejected as impossible for a probability
(c) B1 reason: since A and B are independent, A and B' are also independent, so P(A | B') = P(A)
(c) B1ft awrt 0.390, following through from part (b)
(c) Answer: P(A | B') = P(A) = 0.390 (3 s.f.)