Mathematics 9709/51 October/November 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, October/November 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Two ordinary fair dice, one red and the other blue, are thrown.
Event A is ‘the score on the red die is divisible by 3’.
Event B is ‘the sum of the two scores is at least 9’.
(a) Find P A B. [2]
(b) Hence determine whether or not the events A and B are independent. [2]
Mark scheme
1(a) M1 Complete outcome space or
Red
or listing A and B outcomes
or listing A∩B outcomes
1 2 3 4 5 6
1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
Blue
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12
5 A1 With evidence
P(A∩B) =
36
2
1(b) 1 10 M1 1 10
P(A) × P(B) = × Their × their seen
3 36 3 36
5 5 A1 5 5
≠ so not independent , , P(A) × P(B) and P(A∩B) seen in workings and correct
54 36 54 36
conclusion stated
5
Condone being stated in (a)
36
Alternative method for question 1(b)
P(B|A) = P(B) M1 their 1( a )
5 OE, seen
their P ( A )
P ( A ∩ B ) 36
P(B|A) = =
P (A) 1
3
5 5 A1 5 5
≠ so not independent P(A|B), P(B), , seen in workings and correct conclusion
12 18 12 18
stated
5 10
Condone ≡ being identified in (a)
18 36
2
The probability that a student at a large music college plays in the band is 0.6. For a student who plays in the band, the probability that she also sings in the choir is 0.3. For a student who does not play in the band, the probability that she sings in the choir is x. The probability that a randomly chosen student from the college does not sing in the choir is 0.58.
(a) Find the value of x. [3]
Two students from the college are chosen at random.
(b) Find the probability that both students play in the band and both sing in the choir. [2]
Mark scheme
2(a) 0·6 × 0·7 + 0·4(1 – x) = 0·58 M1 Equation of form 0·6 × a + 0·4 × b = 0·58;
≡ 0·42 + 0·4(1 – x) = 0·58 a = 0·3, 0·7, b = x, (1 – x)
B1 Single correct product seen, condone 0·42, in an equation of
appropriate form
x = 0·6 A1
Alternative method for question 2(a)
0·6 × 0·3 + 0·4x = 0·42 M1 Equation of form 0·6 x a + 0·4 x b = 0·42;
≡ 0·18 + 0·4x = 0·42 a = 0·3, 0·7, b = x, (1 – x)
B1 Single correct product seen, condone 0·18, in an equation of
appropriate form
x = 0·6 A1
3
2(b) ( 0.6 × 0.3)2 M1 (a × b)2, a = 0·6, 0·4 and b = 0·7, 0·3, x, (1–x)
or 0·182, alone.
0.0324 A1
2
Kayla is competing in a throwing event. A throw is counted as a success if the distance achieved is greater than 30 metres. The probability that Kayla will achieve a success on any throw is 0.25.
(a) Find the probability that Kayla takes more than 6 throws to achieve a success. [2]
(b) Find the probability that, for a random sample of 10 throws, Kayla achieves at least 3 successes.
[3]
Mark scheme
3(a) P(X > 6) = 0.756 M1 pn, n = 6, 7 0 < p < 1
729 A1 0·17797…
0.178,
4096
2
3(b) 10 10 1 9 10 2
1 – P(0, 1, 2) = 1 – ( 0.75 + C1 0.25 0.75 + C2 0.25 0.75 ) 8 M1 Binomial term of form 10Cx p x (1 − p )10− x , 0 < p < 1,
any p, x ≠ 0, 10
1 – (0·0563135 + 0·1877117 + 0·2815676) A1 Correct unsimplified expression
0·474 A1 0·474 ⩽ p ⩽ 0·4744
3
The random variable X takes each of the values 1, 2, 3, 4 with probability 41 . Two independent values of X are chosen at random. If the two values of X are the same, the random variable Y takes that value. Otherwise, the value of Y is the larger value of X minus the smaller value of X .
(a) Draw up the probability distribution table for Y . [4]
(b) Find the probability that Y = 2 given that Y is even. [2]
Mark scheme
4(a) B1
y 1 2 3 4 1 2 3 4
7 5 3 1 1 1 1 2 3
prob
16 16 16 16
2 1 2 1 2
3 2 1 3 1
4 3 2 1 4
Probability distribution table with correct scores with at least
one probability, allow extra score values if probability of zero
stated’
B1 One probability (linked with correct score) correct
B1 2 more probs (linked with correct scores) correct
B1 FT 4th prob correct, FT sum of 3 or 4 terms = 1
4
4(b) 5 M1 their P ( 2 )
seen or correct outcome space.
P(2|even) = 16 their P ( 2 ) + their P ( 4 )
6
16
5 A1
or 0·833
6
2
The time in hours that Davin plays on his games machine each day is normally distributed with mean 3.5 and standard deviation 0.9.
(a) Find the probability that on a randomly chosen day Davin plays on his games machine for more than 4.2 hours. [3]
(b) On 90% of days Davin plays on his games machine for more than t hours. Find the value of t.
[3]
(c) Calculate an estimate for the number of days in a year (365 days) on which Davin plays on his games machine for between 2.8 and 4.2 hours. [3]
Mark scheme
5(a) 4.2 − 3.5 M1 Using ± standardisation formula, no σ or σ 2 , continuity
P(X > 4.2) = P( z > )
0.9 correction
= P( z > 0.7778)
1 – 0·7818 M1 Appropriate area Φ , from standardisation formula P(z>…) in
final solution
0·218 A1
3
5(b) z = −1.282 B1 ±1.282 seen (critical value)
t − 3.5 M1 An equation using ±standardisation formula with a z-value,
= −1.282
0.9 condone σ , σ 2 and continuity correction
t = 2.35 A1 AWRT, only dependent on M mark
3
5(c) P(2.8 < X < 4.2) = 1 – 2 × their 5(a) B1 FT FT from their 5(a) < 0.5 or correct
≡ 2(1 – their 5(a)) – 1 Accept unevaluated probability
≡ 2(0·5 – their 5(a)) OE
= 0·5636 Accept 0·564
Number of days = 365 × 0·5636 = 205·7 M1 365 × their p
So, 205 (days) A1 FT Accept 205 or 206, not 205·0 or 206·0 no approximation/
rounding stated
FT must be an integer value
Alternative method for question 5(c)
2.8 − 3.5 4.2 − 3.5 B1 0·5635 < p ⩽ 0·564
P <z<
0.9 0.9
= Φ ( 0.7778) − (1 − Φ0.7778) OE
= 0·7818 – (1 – 0·7818)
= 0·5636
Number of days = 365 × 0·5636 = 205·7 M1 365 × their p
So, 205 (days) A1 FT Accept 205 or 206, not 205·0 or 206·0 no approximation/
rounding stated
FT must be an integer value
3
The times, t minutes, taken by 150 students to complete a particular challenge are summarised in the following cumulative frequency table.
Time taken (t minutes) t ≤ 20 t ≤ 30 t ≤ 40 t ≤ 60 t ≤ 100
Cumulative frequency 12 48 106 134 150
(a) Draw a cumulative frequency graph to illustrate the data. [2]
(b) 24% of the students take k minutes or longer to complete the challenge. Use your graph to estimate the value of k. [2]
(c) Calculate estimates of the mean and the standard deviation of the time taken to complete the challenge. [6]
Mark scheme
6(a) M1 At least 4 points plotted at upper end points, with both scales
linear with at least 3 values indicated
Correct cumulative frequency curve A1 All plotted correctly with curve drawn joined to (0, 0), axes
labelled cumulative frequency, time, minutes
2
6(b) 150 × 0·76 = 114 M1 114 SOI, may be on graph
k = 45 (mins) A1 FT Clear indication that their graph has been used, tolerance ±1mm
2
6(c) Frequencies: 12 36 58 28 16 B1 Correct frequencies seen
10 × 12 + 25 × 36 + 35 × 58 + 50 × 28 + 80 × 16 B1 At least 4 correct midpoints seen and used
Mean =
150
120 + 900 + 2030 + 1400 + 1280 M1 Correct formula with their midpoints (not upper boundary,
150 lower boundary, class width or frequency density).
1 A1
38.2, 38
5
12 × 102 + 36 × 252 + 58 × 352 + 28 × 502 + 16 × 802 M1 Substitute their midpoints and frequencies (condone use of
Variance = − mean 2 cumulative frequency) in correct variance formula, must have
150
‘– their mean2’
1200 + 22500 + 71050 + 70000 + 102400
= − mean 2
150
(Standard deviation = 321.76 ) = 17.9 A1
6
(a) Find the number of different ways in which the 10 letters of the word SHOPKEEPER can be arranged so that all 3 Es are together. [2]
(b) Find the number of different ways in which the 10 letters of the word SHOPKEEPER can be arranged so that the Ps are not next to each other. [4]
(c) Find the probability that a randomly chosen arrangement of the 10 letters of the word
SHOPKEEPER has an E at the beginning and an E at the end. [2]
Four letters are selected from the 10 letters of the word SHOPKEEPER.
(d) Find the number of different selections if the four letters include exactly one P. [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) 8! M1 8! 7!× 8 a!
≡ , where k ∈ , , where a ∈
2! k k 2 (!)
20160 A1
2
7(b) 10! B1 Accept unsimplified
Total number of ways: ( = 302 400 ) (A)
2!3!
9! B1 Accept unsimplified
With Ps together: (= 60 480) (B)
3!
With Ps not together: 302 400 – 60 480 M1 10! 9!
− , m, n integers or (A) – (B) if clearly identified
m n
241 920 A1
Alternative method for question 7(b)
8! B1 k × 8! in numerator, k a positive integer, no ±
3!
B1 m × 3! in denominator, m a positive integer, no ±
9×8 M1 8!
× Their multiplied by 9C2 or 9P2 no additional terms
2 3!
241 920 A1 Exact value, WWW
4
7(c) Number of ways Es at beginning and end M1 8!
Probability =
Total number of ways k ! 1 ⩽ k, l ∈ ℕ ⩽ 3, FT denominator from 7(b) or correct
8! 10!
20160 k !l !
Probability = 2! =
10! 302 400
2!× 3!
1 A1
, 0·0667
15
Alternative method for question 7(c)
3 2 M1 a a −1
Probability = × × a = 3, 2
10 9 10 9
1 A1
, 0·0667
15
Alternative method for question 7(c)
1 1 M1 1 1
Probability = × × 3! × × m!, m = 3, 2
10 9 10 9
1 A1
, 0.0667
15
2
5
7(d) Scenarios: M1 Cx seen alone, 1 ⩽ x ⩽ 4
5
PEEE C0 = 1
5
PEE_ C1 = 5 M1 Summing the number of ways for 3 or 4 correct scenarios (can
5
PE__ C2 = 10 be unsimplified), no incorrect scenarios
5
P___ C3 = 10
Total = 26 A1
3
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On this page, free: the mark scheme for AS & A Level Mathematics 9709/51, October/November 2020 (9709_w20_ms_51), and the question paper (9709_w20_qp_51), to read or download, with each question's marking points written out underneath.
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Mathematics 9709/51 for October/November 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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