Mathematics 9709/51 May/June 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, May/June 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A bag contains 12 marbles, each of a different size. 8 of the marbles are red and 4 of the marbles are blue.
How many different selections of 5 marbles contain at least 4 marbles of the same colour? [4]
Mark scheme
1 RRRRB 8C4 × 4C1 = 280 M1 8
Cx × 4Cy with x + y = 5. x, y both integers, 1 ⩽ x ⩽ 5,
BBBBR 8C1 × 4C4 = 8 0 ⩽ y ⩽ 4 condone 8C1 × 1
RRRRR 8C5 = 56
A1 Two correct outcomes evaluated
M1 Add 2 or 3 identified correct scenarios only (no
additional terms, not probabilities)
[Total =] 344 A1 WWW, only dependent on 2nd M mark
4 SC not all (or no) scenarios identified
B1 280 + 8 + 56 DB1 344
A company produces a particular type of metal rod. The lengths of these rods are normally distributed with mean 25.2 cm and standard deviation 0.4 cm. A random sample of 500 of these rods is chosen.
How many rods in this sample would you expect to have a length that is within 0.5 cm of the mean length? [5]
Mark scheme
2 25.2 − ( 25.5 + 0.50 ) 25.2 − ( 25.2 − 0.50 ) M1 Use of ± Standardisation formula once; no continuity
P < z< correction, σ2, √ σ
0.4 0.4
0.5 0.5
= P − <z<
0.4 0.4
2Φ (1.25 ) − 1
= A1 For AWRT 0.8944 SOI
= 2 × 0.8944 − 1 M1 Appropriate area 2Φ – 1 OE, from final process, must be
probability
0.7888 A1 Accept AWRT 0.789
Number of rods = 0.7888 × 500 B1FT Correct or FT their 4SF (or better) probability, final
= 394 or 395 answer must be positive integer, not 394.0 or 395.0, no
approximation/rounding stated, only 1 answer
5
(a) How many different arrangements are there of the 8 letters in the word RELEASED? [1]
(b) How many different arrangements are there of the 8 letters in the word RELEASED in which the letters LED appear together in that order? [3]
(c) An arrangement of the 8 letters in the word RELEASED is chosen at random.
Find the probability that the letters A and D are not together. [4]
Mark scheme
3(a) 8! B1 NFWW, must be evaluated
3! = 6720
1
3(b) 6! M1 6! 5! x6
_ _ _ L E D _ _ : With LED together: or k ⩾ 1 and no other terms
2! k k
M1 m
, m an integer, m ⩾ 5
2!
360 A1 CAO
3
3(c) Method using _ _ _ A _ D _ _ : *M1 6!
× k seen, k an integer > 0
6! 3!
Arrange the 6 letters RELESE = [= 120]
3!
Multiply by number of ways of placing AD in non-adjacent places *M1 m × n ( n − 1) or m × nC2 or m × n P2 , n = 6, 7 or 8, m an
= their 120 × 7P2 [= 5040]
integer > 0
their 5040 DM1 Denominator = their (a) or correct, dependent on at least
[Probability =] one M mark already gained.
their 6720
5040 3 A1
or or 0.75
6720 4
Alternative method for Question 3(c)
Method using ‘Total arrangements – Arrangements with A and D *M1 Their 6720 – k, k a positive integer
together’:
7!×2 *M1 7!× k
Their 6720 – [= 5040] ( m −) , k = 1, 2
3! 3!
their 5040 DM1 With denominator = their (a) or correct, dependent on at
[Probability =] least one M mark already gained.
their 6720
5040 3 A1
or or 0.75
6720 4
Alternative method for Question 3(c)
Method using ‘1 – Probability of arrangements with A and D together’: *M1 7!× k
, k = 1, 2
7!×2 3!
[= 1680]
3!
their 1680 *M1 With denominator = their (a) or correct
[Probability =]
their 6720
their 1680 DM1 1 – m, 0 < m < 1 , dependent on at least one M mark
1– already gained
their 6720
5040 3 A1
or or 0.75
6720 4
4
To gain a place at a science college, students first have to pass a written test and then a practical test.
Each student is allowed a maximum of two attempts at the written test. A student is only allowed a second attempt if they fail the first attempt. No student is allowed more than one attempt at the practical test. If a student fails both attempts at the written test, then they cannot attempt the practical test.
The probability that a student will pass the written test at the first attempt is 0.8. If a student fails the first attempt at the written test, the probability that they will pass at the second attempt is 0.6. The probability that a student will pass the practical test is always 0.3.
(a) Draw a tree diagram to represent this information, showing the probabilities on the branches.
[3]
(b) Find the probability that a randomly chosen student will succeed in gaining a place at the college.
[2]
(c) Find the probability that a randomly chosen student passes the written test at the first attempt given that the student succeeds in gaining a place at the college. [2]
Mark scheme
4(a) B1 Fully correct labelled tree diagram for each pair of
branches clearly identifying written and practical, pass
and fail for each intersection (no additional branches)
B1 ‘One written test’ branch all probabilities (or %) correct
B1 ‘Two written tests’ branch all probabilities (or %)
correct, condone additional branches after W2F with
probabilities 1 for PF and 0 for PP
3
4(b) [P(W1P) × P(PP) + P(W1F) × P(W2P) × P(PP)] M1 Consistent with their tree diagram or correct
0.8 × 0.3 + 0.2 × 0.6 × 0.3
69 A1
0.276 or
250
2
4(c) P (W 1 ∩ Practical ) 0.8 × 0.3 0.24 M1 Correct expression or FT their (b)
P (W 1 P ) = = =
P ( getting place ) their ( b ) 0.276
20 A1
or 0.87[0]
23
2
The times taken by 200 players to solve a computer puzzle are summarised in the following table.
Time (t seconds) 0 ≤ t < 10 10 ≤ t < 20 20 ≤ t < 40 40 ≤ t < 60 60 ≤ t < 100
Number of players 16 54 78 32 20
(a) Draw a histogram to represent this information. [4]
(b) Calculate an estimate of the mean time taken by these 200 players. [2]
(c) Find the greatest possible value of the interquartile range of these times. [2]
Mark scheme
5(a) M1 At least 4 frequency densities calculated, accept
unsimplified.
Class width 10 10 20 20 40
May be read from graph using their scale, 3SF or correct
Frequency Density 1.6 5.4 3.9 1.6 0.5
A1 All heights correct on graph
B1 Bar ends at 0, 10, 20 …, etc. with a horizontal linear
scale with at least 3 values indicated,
0 ⩽ horizontal axis ⩽ 100
B1 Axes labelled: Frequency density (fd), time (t) and
seconds. Linear vertical scale, with at least 3 values
indicated 0 ⩽ vertical axis ⩽ 5.4
4
5(b) 16 × 5 + 54 × 15 + 78 × 30 + 32 × 50 + 20 × 80 M1 Uses at least 4 midpoint attempts (e.g. 5 ± 0.5).
Mean = Accept unsimplified expression, denominator either
200
correct or their Σfrequencies
80 + 810 + 2340 + 1600 + 1600
=
200
6430 3 A1 Accept 32.2
200 = 32 20 or 32.15
2
5(c) A value in correct UQ (40–60) – a value in correct LQ (10–20) M1
Greatest possible value is 60 – 10 = 50 A1 Condone 49. 9
2
In Questa, 60% of the adults travel to work by car.
(a) A random sample of 12 adults from Questa is taken.
Find the probability that the number who travel to work by car is less than 10. [3]
(b) A random sample of 150 adults from Questa is taken.
Use an approximation to find the probability that the number who travel to work by car is less than 81. [5]
(c) Justify the use of your approximation in part (b). [1]
Mark scheme
6(a) 1 – P(10, 11, 12) = 1 – (12C10 0.610 0.42 + 12C11 0.611 0.41 + 12C12 0.612 0.40) M1 One term: 12Cx px (1 – p)12-x for 0 < x < 12, any p allowed.
[= 1 – (0.063852 + 0.017414 + 0.0021768)]
A1 Correct unsimplified expression, or better.
[1 – 0.083443] = 0.917 A1 AWRT
Alternative method for Question 6(a)
P (0,1,2,3,4,5,6,7,8,9) = 12C00.60 0.412 + 12C1 0.61 0.411+ ………….12C9 0.69 M1 One term: 12Cx px (1 – p)12-x for 0 < x < 12, any p allowed.
0.43
[= 0.000016777 + 0.00030199 + 0.0024914 + 0.012457 + 0.042043 + A1 Correct unsimplified expression with at least the first
0.10090 + 0.17658 + 0.22703 + 0.21284 + 0.14189] two and last terms
0.917 A1 WWW, AWRT
3
6(b) [Mean =] 0.6 × 150 [= 90]; B1 Correct mean and variance. Accept evaluated or
[Variance =] 0.6 × 150 × 0.4 [= 36] unsimplified
80.5 − 90 M1 Substituting their mean and variance into
P ( X < 81) = P Z < ±standardisation formula (with a numerical value for
6
80.5), allow σ2, √ σ, but not µ ± 0.5
M1 Using continuity correction 80.5 or 81.5
Φ ( −1.5833) = 1 − 0.9433 M1 Appropriate area Φ, from final process, must be
probability
0.0567 A1 AWRT
5
6(c) np = 90, nq = 60 both greater than 5 B1 At least nq evaluated and statement >5 required
1
Sharma knows that she has 3 tins of carrots, 2 tins of peas and 2 tins of sweetcorn in her cupboard.
All the tins are the same shape and size, but the labels have all been removed, so Sharma does not know what each tin contains.
Sharma wants carrots for her meal, and she starts opening the tins one at a time, chosen randomly, until she opens a tin of carrots. The random variable X is the number of tins that she needs to open.
6.
(a) Show that P X = 3 = 35 [2]
(b) Draw up the probability distribution table for X . [4]
(c) Find Var X . [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) 4 3 3 M1 m n o 1
P ( X = 3) = × × × × used throughout. condone use of
7 6 5 7 6 5 2
6 A1 AG.
35 The fractions must be identified, e.g. P(NC, NC, C), may
be seen in a tree diagram.
2
7(b) B1 Table with x values and at least one probability
x 1 2 3 4 5
Condone any additional x values if probability stated as
15 10 6 3 1 0.
p
35 35 35 35 35
B1 One correct probability other than X = 3 linked to the
correct outcome
B1 Two further correct probabilities other than X = 3 seen
linked to the correct outcome
B1FT All probabilities correct, or at least 4 probabilities
summing to 1
4
7(c) 15 10 6 3 1 M1 At least 4 correct terms FT their values in (a) with
[E(X) = 1× + 2× + 3× + 4× + 5× ] probabilities summing to 1
35 35 35 35 35
May be implied by use in Variance, accept unsimplified
expression.
15 + 20 + 18 + 12 + 5 70
E(X ) = = 35 = 2
35
12 × 15 + 22 × 10 + 32 × 6 + 42 × 3 + 52 × 1 2 M1 Appropriate variance formula using their (E(X))2.
Var ( X ) = − 2 = FT their table accept probabilities not summing to 1.
35
15 + 40 + 54 + 48 + 25 2
−2
35
182 6 A1 N.B. If method FT for M marks from their incorrect (b),
= 35 − 4 = 5 expressions for E(X) and Var(X) must be seen
unsimplified with all probabilities <1
3
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Mathematics 9709/51 for May/June 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
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