Statistics: Probability Depth - Worksheets, Questions and Revision

12 original exam-style questions - 4 pages of questions with a full mark scheme - free printable PDF.

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A-Level · Statistics

S7 Statistics: Probability Depth

EDEXCEL 9MA0 · Calculator allowed · about 130 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A-Level Statistics: Probability Depth Practice

Original material written for Revision Library, styled on Edexcel AS/A-level Mathematics (9MA0) Statistics content.

This pack extends core probability skills to full A-level depth: Venn diagrams and set notation for two and three events, tree diagrams for dependent and independent trials, conditional probability, testing and proving independence algebraically, and probability models involving an unknown to be found from a given equation, including quadratics arising from probability statements. All contexts and figures are original.

At a sixth-form college in Leeds, A is the event that a randomly chosen student studies History and B is the event that a randomly chosen student studies Psychology. It is known that P(A) = 0.55, P(B) = 0.35 and P(A and B) = 0.20.

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

(b)Find P(not A and not B).(1)

(Total for Question 1 is 3 marks)

For two events C and D, P(C) = 0.4, P(D) = 0.3 and P(C or D) = 0.58.

(a)Determine, showing your working, whether C and D are independent.(3)

(b)State the value of P(D | C).(1)

(Total for Question 2 is 4 marks)

At a college with 200 students in total, E is the event that a randomly chosen student studies Economics and B is the event that a randomly chosen student studies Business. It is known that 84 students study Economics, 90 students study Business, and 37 students study both Economics and Business.

(a)Find the number of students who study neither Economics nor Business.(2)

(b)Find P(not E and B), the probability that a randomly chosen student studies Business but not Economics.(2)

(c)Find P(E | B).(2)

(Total for Question 3 is 6 marks)

A gym in Cardiff has 120 members. Y is the event that a randomly chosen member attends yoga classes and P is the event that a randomly chosen member attends Pilates classes. Of the 120 members: 28 attend yoga only, 34 attend Pilates only, 22 attend both yoga and Pilates, and 36 attend neither.

(a)Find P(Y).(1)

(b)Find P(Y | P).(2)

(c)Find P(P | not Y).(2)

(d)Determine, showing your working, whether Y and P are independent.(2)

(Total for Question 4 is 7 marks)

A bag contains 9 counters: 5 red and 4 blue. Two counters are drawn at random from the bag, one after another, without replacement.

(a)Find the probability that both counters drawn are red.(2)

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

(c)Given that the first counter drawn is blue, find the probability that the second counter drawn is red.(2)

(Total for Question 5 is 7 marks)

For events F and G, P(F) = 0.6, P(G|F) = 0.25 and P(G|not F) = p, where p is a constant. It is given that P(G) = 0.34.

(a)Find the value of p.(3)

(b)Find P(F | G).(3)

(c)Find P(not F | not G).(3)

(Total for Question 6 is 9 marks)

Events J and K are mutually exclusive, with P(J) = 1/3 and P(K) = 1/4.

(a)Write down P(J and K).(1)

(b)Find P(J or K).(1)

(c)L is the event that neither J nor K occurs. Find P(L).(2)

(d)Explain, with reference to your answer to part (a), why J and K cannot be independent.(3)

(Total for Question 7 is 7 marks)

A machine in a factory produces electronic components. Each component produced is faulty, independently of all other components, with probability 0.04. A random sample of 3 components is checked.

(a)Find the probability that exactly one of the three components is faulty.(3)

(b)Find the probability that at least one of the three components is faulty.(3)

(c)State the probability that none of the three components is faulty.(1)

(d)Given that at least one of the three components is faulty, find the probability that exactly one of the three components is faulty.(2)

(Total for Question 8 is 9 marks)

Events M and N are independent, with P(M) = x and P(N) = x + 0.2, where 0 < x < 0.8. It is given that P(M or N) = 0.6.

(a)Show that 5x² - 9x + 2 = 0.(4)

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

(c)Find P(M and N).(2)

(d)Determine whether M and not N are independent, justifying your answer.(2)

(Total for Question 9 is 11 marks)

In a survey of 94 students at a sixth-form college in Bristol, D is the event that a student takes part in Drama, C is the event that a student takes part in Choir, and S is the event that a student takes part in Sport. Using a Venn diagram of the survey results: the number of students in D only is 2x, in C only is 3x, in S only is x, in D and C only (not S) is x, in D and S only (not C) is 7, in C and S only (not D) is 10, in all three of D, C and S is 6, and in none of D, C or S is 15.

Figure (to be drawn): Three-circle Venn diagram (D, C, S) inside a rectangle, with region counts 2x, 3x, x, x, 7, 10, 6 and 15 outside all three circles, as described in the question stem.

(a)Form an equation in x and hence find the value of x.(3)

(b)Using x = 8, find P(D).(2)

(c)Find P(S | C).(2)

(d)Determine, showing your working, whether D and S are independent.(2)

(Total for Question 10 is 9 marks)

A screening test is used to detect a medical condition in a large population, in which 2% of people have the condition. Let H be the event that a randomly chosen person has the condition, and T be the event that the test gives a positive result. If a person has the condition, the test gives a positive result with probability 0.97. If a person does not have the condition, the test still incorrectly gives a positive result with probability 0.04.

Figure (to be drawn): Two-stage tree diagram: first branch H (0.02) / not H (0.98); second branch from each, T / not T with the conditional probabilities given in the question.

(a)State one assumption needed for the tree diagram model above to be valid.(1)

(b)Find P(T).(3)

(c)Find the probability that a randomly selected person does not have the condition, given that their test result is positive.(4)

(d)Comment on the usefulness of this test as a diagnostic tool, with reference to your answer to part (c).(3)

(Total for Question 11 is 11 marks)

A bag contains n red counters and 8 blue counters, where n is a positive integer. Two counters are drawn at random from the bag, one after another, without replacement.

(a)Show that P(both counters drawn are red) = n(n-1) / [(n+8)(n+7)].(2)

(b)Given that P(both counters drawn are red) = 5/17, show that 3n² - 23n - 70 = 0.(4)

(c)Hence find the value of n, explaining why the other solution of the equation is rejected.(2)

(d)Using n = 10, given that the first counter drawn is blue, find the probability that the second counter drawn is red.(1)

(e)Using n = 10, find the probability that exactly one of the two counters drawn is red.(3)

(Total for Question 12 is 12 marks)

Mark scheme · S7 Statistics: Probability Depth

Question 1

(a) M1 P(A) + P(B) - P(A and B)
(a) A1 cao 0.70
(a) Answer: P(A or B) = 0.70
(b) B1ft 1 - P(A or B), following through from part (a); cao 0.30
(b) Answer: P(not A and not B) = 0.30

Question 2

(a) M1 P(C and D) = P(C) + P(D) - P(C or D)
(a) A1 P(C and D) = 0.12, and P(C) x P(D) = 0.4 x 0.3 = 0.12
(a) A1 cso: since P(C and D) = P(C) x P(D) (both equal 0.12), C and D are independent
(a) Answer: Independent, since P(C and D) = 0.12 = P(C) x P(D)
(b) B1ft cao 0.3, using independence from part (a) so P(D | C) = P(D)
(b) Answer: P(D | C) = 0.3

Question 3

(a) M1 n(E or B) = 84 + 90 - 37
(a) A1 cao 63, from 200 - 137
(a) Answer: 63 students study neither Economics nor Business
(b) M1 n(B) - n(E and B) = 90 - 37
(b) A1 cao 53/200 (= 0.265)
(b) Answer: P(not E and B) = 53/200 = 0.265
(c) M1 n(E and B) / n(B), oe
(c) A1 cao 37/90 (awrt 0.411)
(c) Answer: P(E | B) = 37/90 (awrt 0.411)

Question 4

(a) B1 cao 50/120 (= 5/12, awrt 0.417)
(a) Answer: P(Y) = 5/12 (awrt 0.417)
(b) M1 22/(34+22), oe
(b) A1 cao 11/28 (awrt 0.393)
(b) Answer: P(Y | P) = 11/28 (awrt 0.393)
(c) M1 identify n(not Y) = 120 - 50 = 70 and n(P and not Y) = 34
(c) A1 cao 17/35 (awrt 0.486)
(c) Answer: P(P | not Y) = 17/35 (awrt 0.486)
(d) M1 compute P(Y) x P(P) = (5/12) x (7/15), using P(P) = 56/120 = 7/15
(d) A1 cso: P(Y) x P(P) = 7/36 (awrt 0.194), which does not equal P(Y and P) = 11/60 (awrt 0.183), so Y and P are not independent
(d) Answer: Not independent, since P(Y)P(P) = 7/36 (awrt 0.194) not equal to P(Y and P) = 11/60 (awrt 0.183)

Question 5

(a) M1 (5/9) x (4/8), oe
(a) A1 cao 5/18 (awrt 0.278)
(a) Answer: P(both red) = 5/18 (awrt 0.278)
(b) M1 P(red then blue) = (5/9) x (4/8), oe
(b) M1 P(blue then red) = (4/9) x (5/8), oe, and adding the two paths
(b) A1 cao 5/9 (awrt 0.556)
(b) Answer: P(exactly one red) = 5/9 (awrt 0.556)
(c) M1 identify the remaining counters after one blue is removed: 5 red and 3 blue (8 left)
(c) A1 cao 5/8 (= 0.625)
(c) Answer: P(second red | first blue) = 5/8 (= 0.625)

Question 6

(a) M1 P(F and G) = P(F) x P(G|F) = 0.6 x 0.25 = 0.15
(a) M1 use P(G) = P(F and G) + P(not F) x p, i.e. 0.34 = 0.15 + 0.4p
(a) A1 cao p = 0.475
(a) Answer: p = 0.475
(b) M1 identify P(F and G) = 0.15 (ft from a)
(b) M1 divide by P(G) = 0.34
(b) A1 cao 15/34 (awrt 0.441)
(b) Answer: P(F | G) = 15/34 (awrt 0.441)
(c) M1 P(F or G) = P(F) + P(G) - P(F and G) = 0.6 + 0.34 - 0.15 = 0.79, then P(not F and not G) = 1 - 0.79
(c) M1 P(not G) = 1 - 0.34 = 0.66, then divide: 0.21/0.66
(c) A1 cao 7/22 (awrt 0.318)
(c) Answer: P(not F | not G) = 7/22 (awrt 0.318)

Question 7

(a) B1 cao 0
(a) Answer: P(J and K) = 0
(b) B1ft cao 7/12, using P(J or K) = P(J) + P(K) since P(J and K) = 0 (ft from a)
(b) Answer: P(J or K) = 7/12
(c) M1ft 1 - P(J or K), following through from part (b)
(c) A1 cao 5/12
(c) Answer: P(L) = 5/12
(d) B1 states the condition for independence: P(J and K) = P(J) x P(K), oe
(d) B1 calculates P(J) x P(K) = (1/3)(1/4) = 1/12
(d) B1 correct conclusion: since P(J and K) = 0 (from part a) and 0 does not equal 1/12, and P(J), P(K) are both nonzero, J and K cannot be independent
(d) Answer: Since P(J and K) = 0 but P(J) x P(K) = 1/12, and these are not equal, J and K cannot be independent

Question 8

(a) M1 identify 3 ways of choosing which component is faulty, e.g. 3 x 0.04 x 0.96²
(a) M1 correct substitution: 3 x 0.04 x 0.9216
(a) A1 awrt 0.111
(a) Answer: P(exactly one faulty) = 0.111 (3 s.f.)
(b) M1 P(none faulty) = 0.96³
(b) M1 1 - P(none faulty)
(b) A1 awrt 0.115
(b) Answer: P(at least one faulty) = 0.115 (3 s.f.)
(c) B1ft cao awrt 0.885, following through from part (b) as 1 - their answer
(c) Answer: P(none faulty) = 0.885 (3 s.f.)
(d) M1ft their (a) / their (b), using the fact that 'exactly one' implies 'at least one'
(d) A1 awrt 0.959
(d) Answer: P(exactly one | at least one) = awrt 0.959

Question 9

(a) M1 use P(M or N) = P(M) + P(N) - P(M)P(N), using independence
(a) M1 correct substitution: x + (x + 0.2) - x(x + 0.2) = 0.6
(a) dM1 expand and rearrange into a three-term quadratic set to zero
(a) A1 cso: 5x² - 9x + 2 = 0, fully shown
(a) Answer: 5x² - 9x + 2 = 0 (shown)
(b) M1 solve 5x² - 9x + 2 = 0 using the quadratic formula or a calculator
(b) A1 awrt x = 0.260 (other root awrt 1.540)
(b) B1 reason: the other root (awrt 1.540) is rejected as it does not satisfy 0 < x < 0.8 (and exceeds 1, which is impossible for a probability), oe
(b) Answer: x = 0.260 (3 s.f.); other root x = 1.540 rejected as it lies outside 0 < x < 0.8
(c) M1ft x(x + 0.2), using their value of x from part (b)
(c) A1 awrt 0.119
(c) Answer: P(M and N) = awrt 0.119
(d) B1 states the result that if two events are independent, then one event and the complement of the other event are also independent, oe
(d) B1ft correct conclusion: since M and N are independent, M and not N are also independent
(d) Answer: M and not N are independent, since M and N are independent

Question 10

(a) M1 sum all 8 regions and set equal to 94
(a) A1 correct equation 7x + 38 = 94, oe
(a) A1 cao x = 8
(a) Answer: x = 8
(b) M1ft n(D) = 16 + 8 + 7 + 6, using their x
(b) A1 cao 37/94 (awrt 0.394)
(b) Answer: P(D) = 37/94 (awrt 0.394)
(c) M1ft n(C and S)/n(C) = (10+6)/(24+8+10+6), using their x
(c) A1 cao 1/3 (awrt 0.333)
(c) Answer: P(S | C) = 1/3 (awrt 0.333)
(d) M1ft compute P(D) x P(S) = (37/94) x (31/94), using their x
(d) A1 cso: P(D) x P(S) = awrt 0.130, which does not equal P(D and S) = 13/94 (awrt 0.138), so D and S are not independent
(d) Answer: Not independent, since P(D)P(S) = awrt 0.130 not equal to P(D and S) = 13/94 (awrt 0.138)

Question 11

(a) B1 e.g. whether one person has the condition (or tests positive) is independent of whether any other person has the condition (or tests positive), oe; or the given sensitivity and false-positive rate apply equally to every person tested, oe
(a) Answer: For example: the test result for one person is independent of the test result for any other person
(b) M1 P(H) x P(T|H) = 0.02 x 0.97
(b) M1 P(not H) x P(T|not H) = 0.98 x 0.04, and add the two products
(b) A1 cao 0.0586
(b) Answer: P(T) = 0.0586
(c) M1 identify P(not H and T) = 0.0392
(c) M1ft divide by P(T), following through from part (b)
(c) A1 0.0392/0.0586 or exact fraction 196/293
(c) A1 awrt 0.669
(c) Answer: P(not H | T) = 196/293 (awrt 0.669)
(d) B1 despite a positive result, there is still a probability of about 0.669 (roughly two out of three) that the person does not have the condition, oe
(d) B1 this is because the condition is rare (only 2% prevalence), so false positives from the large healthy population outnumber true positives from the small group who have the condition, oe
(d) B1 conclusion: a single positive result alone is not sufficient to diagnose the condition; further/confirmatory testing would be needed, oe
(d) Answer: The test is not very reliable on its own for a rare condition: a positive result still leaves about a 0.669 chance the person is unaffected, because far more healthy people are tested than affected people, so their false positives dominate; confirmatory testing is needed