Mathematics 9709/53 May/June 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, May/June 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
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 B1
probability 0.
2
1(b) P (C ∩ E) 0.2 × 0.6 M1
P(C|E) = =
P(E) 0.2 × 0.6 + 0.45 × 0.1 + 0.35 × 1
Summing three appropriate 2-factor probabilities M1
0.12 A1
0.515
12 A1
0.233 or
515
4
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]
In a certain town, the time, X hours, for which people watch television in a week has a normal distribution with mean 15.8 hours and standard deviation 4.2 hours.
(a) Find the probability that a randomly chosen person from this town watches television for less than 21 hours in a week. [2]
(b) Find the value of k such that P X < k = 0.75. [3]
Mark scheme
3(a) 21 − 15.8 M1
P(X < 21) = P z < = Φ(1.238)
4.2
0.892 A1
2
3(b) z = ±0.674 B1
k − 15.8 M1
= 0.674
4.2
18.6 A1
3
A fair four-sided spinner has edges numbered 1, 2, 2, 3. A fair three-sided spinner has edges numbered
−2, −1, 1. Each spinner is spun and the number on the edge on which it comes to rest is noted. The random variable X is the sum of the two numbers that have been noted.
(a) Draw up the probability distribution table for X . [3]
(b) Find Var X . [3]
Mark scheme
4(a)
–1 0 0 1
0 1 1 2
2 3 3 4
x –1 0 1 2 3 4
Probability 1 3 3 2 2 1
12 12 12 12 12 12
Probability distribution table with correct scores with at least one probability B1
At least 4 probabilities correct B1
All probabilities correct B1
3
4(b) −1 + 0 + 3 + 4 + 6 + 4 16 4 B1
E(X) = = =
12 12 3
1 + 0 + 3 + 8 + 18 + 16 4
2 M1
Var(X) = −
12 3
37 A1
(= 2.06)
18
3
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 B1
1
=4
4
1
5(b) 9 B1
( = 0.141)
64
1
5(c) 3
5 M1
P(X < 6) = 1 –
4
(FT their probability/mean from part (a))
0.763 A1
2
5(d) Mean = 80 × 0.25 = 20 M1
Var = 80 × 0.25 × 0.75 = 15
25.5 − 20 M1
P(more than 25) = P z >
15
P(z > 1.42) M1
1 – 0.9222 M1
0.0778 A1
5
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 = [$]35 000 B1
UQ = [$]52 000
IQR = [$]17 000 B1 FT
(FT if 49000 ≤ UQ ≤ 53000 − 32000 ≤ LQ ≤ 41000 )
3
6(c) Sum of given 11 numbers is 433 000 M1
Sum of 12 numbers, including new = 38 500 × 12 = 462 000 M1
Difference = new salary = [$]29 000 A1
3
(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]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) 9! B1
= 90 720
2!2!
1
7(b) 6! M1
2!
360 A1
2
7(c) 8! M1
2 Es together = ( = 20160 )
2!
Es not together = 90720 – 20160 = 70560 M1
70560 M1
Probability =
90720
7 A1
or 0.778
9
Alternative method for question 7(c)
_^_^_^_^_^_^_^_
7! 8 × 7
× = 70560
2! 2
7! × k in numerator, k integer ⩾ 1, denominator ⩾ 1 M1
Multiplying by 8C2 OE M1
70560 M1
Probability =
90720
7 A1
or 0.778
9
4
7(d) Scenarios are: M1
E L _ _ _ 5C3 10
E E L _ _ 5C2 10
E _ _ _ _ 5C4 5
E E _ _ _ 5C3 10
Summing the number of ways for 3 or 4 correct scenarios M1
Total = 35 A1
3
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Mathematics 9709/53 for May/June 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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