Cambridge A Level Mathematics 9709 — 2020 May/June Paper 5 · Variant 3

1 Juan goes to college each day by any one of car or bus or walking. The probability that he goes by car is 0.2, the probability that he goes by bus is 0.45 and the probability that he walks is 0.35. When Juan goes by car, the probability that he arrives early is 0.6. When he goes by bus, the probability that he arrives early is 0.1. When he walks he always arrives early. (a) Draw a fully labelled tree diagram to represent this information. [2] (b) Find the probability that Juan goes to college by car given that he arrives early. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) Fully correct labelled tree for method of transport with correct probabilities. B1 Fully correct labelled branches with correct probabilities for lateness with either 1 branch after W or 2 branches with the probability 0. B1 2 1(b) P(C|E) = ( ) ( ) P C E 0.2 0.6 P E 0.2 0.6 0.45 0.1 0.35 1 ∩ × = × + × + × M1 Summing three appropriate 2-factor probabilities M1 0.12 0.515 A1 0.233 or 12 515 A1 4

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2 In a certain large college, 22% of students own a car. (a) 3 students from the college are chosen at random. Find the probability that all 3 students own a car. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) 16 students from the college are chosen at random. Find the probability that the number of these students who own a car is at least 2 and at most 4. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) 1 2(b) P(2, 3, 4) = 16C2 2 14 16 3 13 16 4 12 3 4 0.22 0.78 0.22 0.78 0.22 0.78 C C + + M1 0.179205 + 0.235877 + 0.216221 A1 0.631 A1 3

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5 A pair of fair coins is thrown repeatedly until a pair of tails is obtained. The random variable X denotes the number of throws required to obtain a pair of tails. (a) Find the expected value of X. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the probability that exactly 3 throws are required to obtain a pair of tails. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the probability that fewer than 6 throws are required to obtain a pair of tails. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ On a different occasion, a pair of fair coins is thrown 80 times. (d) Use an approximation to find the probability that a pair of tails is obtained more than 25 times. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) 1 4 1 4 = B1 1 5(b) ( ) 9 0.141 64 = B1 1 5(c) P(X < 6) = 1 – 5 3 4       (FT their probability/mean from part (a)) M1 0.763 A1 2 5(d) Mean = 80 × 0.25 = 20 Var = 80 × 0.25 × 0.75 = 15 M1 P(more than 25) = P 25.5 20 15 z −   >     M1 P(z > 1.42) M1 1 – 0.9222 M1 0.0778 A1 5

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6 The annual salaries, in thousands of dollars, for 11 employees at each of two companies A and B are shown below. Company A 30 32 35 41 41 42 47 49 52 53 64 Company B 26 47 30 52 41 38 35 42 49 31 42 (a) Represent the data by drawing a back-to-back stem-and-leaf diagram with company A on the left-hand side of the diagram. [4] (b) Find the median and the interquartile range of the salaries of the employees in company A. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ A new employee joins company B. The mean salary of the 12 employees is now $38 500. (c) Find the salary of the new employee. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) A B 2 6 5 2 0 3 0 1 5 8 9 7 2 1 1 4 1 2 2 7 9 3 2 5 2 4 6 KEY 1 | 4 | 2 means $41 000 for A and $42 000 for B Correct stem B1 Correct A on LHS B1 Correct B on same diagram B1 Correct key for their diagram, both companies identified and correct units B1 4 6(b) Median = [$]42 000 B1 LQ = [$]35 000 UQ = [$]52 000 B1 IQR = [$]17 000 (FT if 49000 UQ 53000 32000 LQ 41000 − ≤ ≤ ≤ ≤ ) B1 FT 3 Question Answer Marks 6(c) Sum of given 11 numbers is 433 000 M1 Sum of 12 numbers, including new = 38 500 × 12 = 462 000 M1 Difference = new salary = [$]29 000 A1 3

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7 (a) Find the number of different possible arrangements of the 9 letters in the word CELESTIAL. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the number of different arrangements of the 9 letters in the word CELESTIAL in which the first letter is C, the fifth letter is T and the last letter is E. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the probability that a randomly chosen arrangement of the 9 letters in the word CELESTIAL does not have the two Es together. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 5 letters are selected at random from the 9 letters in the word CELESTIAL. (d) Find the number of different selections if the 5 letters include at least one E and at most one L. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) 9! 2!2!= 90 720 B1 1 7(b) 6! 2! M1 360 A1 2 Question Answer Marks 7(c) 2 Es together = ( ) 8! 20160 2! = M1 Es not together = 90720 – 20160 = 70560 M1 Probability = 70560 90720 M1 7 or 0.778 9 A1 Alternative method for question 7(c) _ ^ _ ^ _ ^ _ ^ _ ^ _ ^ _ ^ _ 7! 8 7 2! 2 × × = 70560 7! × k in numerator, k integer ⩾ 1, denominator ⩾ 1 M1 Multiplying by 8C2 OE M1 Probability = 70560 90720 M1 7 or 0.778 9 A1 4 Question Answer Marks 7(d) Scenarios are: E L _ _ _ 5C3 10 E E L _ _ 5C2 10 E _ _ _ _ 5C4 5 E E _ _ _ 5C3 10 M1 Summing the number of ways for 3 or 4 correct scenarios M1 Total = 35 A1 3

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