Mathematics 9709/51 October/November 2024: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 5 Probability & Statistics 1, October/November 2024. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
Nicola throws an ordinary fair six-sided dice. The random variable X is the number of throws that she takes to obtain a 6.
(a) Find P (X 1 8) . [2]
(b) Find the probability that Nicola obtains a 6 for the second time on her 8th throw. [2]
Mark scheme
1(a) Method 1
7 M1 5 1
5 1 – bd , b = , d = 7, 8.
[P(X < 8) =] 1 −
6 6 6
5 1
1 – ce – (1 – c) × ce – 1, c = , e = 8,9.
6 6
= 0.721 A1 201811
0.720918… .
279936
201811
If M0 scored, SC B1 for 0.7209… or only.
279936
Method 2
M1 a + ba + b2 a + b3a + b4a + b5a + b6a +b7 a .
2 3 4
1 5 1 5 1 5 1 5 1
+ + + + +
6 6 6 6 6 6 6 6 6 1 5
[P(X < 8) =] 5 6
a = , a + b = 1.
5 1 5 1 6 6
+
6 6 6 6
= 0.721 A1 201811
0.720918… .
279936
201811
If M0 scored, SC B1 for 0.7209… or only.
279936
2
1(b) 6 2 M1 6 2
5 1 5 1
× ×7 × × d d an integer ≥ 1, no inappropriate addition.
6 6 6 6
0.0651 A1 0.0651 ⩽ p < 0.06512.
2
The random variable X takes the values -2, -1, 0, 2, 3. It is given that P(X = x) = k (x 2 + 2) , where k is a positive constant.
(a) Draw up the probability distribution table for X, giving the probabilities as numerical fractions.
[3]
(b) Find the value of Var( X ). [3]
Mark scheme
2(a) [Probs 6k , 3k , 2k , 6k , 11k so 28k = 1, ] B1 k must be identified
1
k=
28
M1 Table with correct outcomes and 2 correct probabilities.
x –2 –1 0 2 3 FT substituting their k correctly into formula, with 0 < p < 1.
No additional x values unless probability 0.
P(X = x) 6 3 2 6 11
6k
28 28 28 28 28 Condone in terms of k of the form or 6k.
28
0.2143 0.1071 0.07143 0.2143 0.3929
A1 Fully correct.
Decimal answers to at least 3 sig figures, condone not
summing exactly to 1.
3
2(b) 6 3 2 6 11 M1 Accept unsimplified expression. May be calculated in the
E(X) = −2 × + −1× + 0 × + 2 × + 3 × variance. FT their table with 5 probabilities 0 < p < 1 that
28 28 28 28 28 sum to 1.
( −12 − 3 + 12 + 33) =
1 15
28 14
6 × ( −2 ) + 3 × ( −1) + 6 × 22 + 11× 32 M1 Appropriate variance formula using their (E(X))2 value. FT
2 2
Var(X) = their table with at least 4 probabilities 0 < p < 1, that may not
28 sum to 1.
2
15
−their
14
41 A1 825
= 4.21, 4 Condone .
196 196
If one or both M marks not awarded, SC B1 for correct
answer WWW.
3
The time taken, in minutes, to walk to school was recorded for 200 pupils at a certain school. These times are summarised in the following table.
Time taken t G 15 t G 25 t G 30 t G 40 t G 50 t G 70
(t minutes)
Cumulative
18 46 88 140 176 200 frequency
(a) Draw a cumulative frequency graph to illustrate the data. [2]
(b) Use your graph to estimate the median and the interquartile range of the data. [3]
(c) Calculate an estimate for the mean value of the times taken by the 200 pupils to walk to school.
[3]
Mark scheme
3(a) M1 At least 4 points plotted within tolerance at upper bounds.
Linear cf scale 0 ⩽ cf ⩽ 200 and linear time scale 0 ⩽ t ⩽ 70,
with at least 3 values identified on each.
Minimum scale uses at least ½ the grid.
A1 All points plotted correctly. Curve drawn and joined to (0, 0).
Axes labelled cumulative frequency (cf), time (t) and minutes
(min) – or a suitable title.
2
3(b) Median = 33 B1 FT Must be identified.
Evidence of use of graph must be seen.
Strict FT ± ½ square on time axis.
[IQR = ] 42 – 26 M1 41 ⩽ UQ ⩽ 43 − 25 < LQ ⩽ 27 .
If outside of range FT ± ½ square on time axis.
16 A1 FT
3
3(c) B1 At least 5 correct midpoints
Midpoint 7.5 20 27.5 35 45 60 or
5 correct frequencies seen.
Frequency 18 28 42 52 36 24
18 × 7.5 + 28 × 20 + 42 × 27.5 + 52 × 35 + 36 × 45 + 24 × 60 M1 Correct mean formula using their 6 midpoints (must be
Mean = within class, not upper bound, not lower bound) condone 1
200
error and their 6 frequencies (not cumulative frequencies).
13 A1 673
= 33.65 , 33 Accept 33.7, not .
20 20
3
Rahul has two bags, X and Y. Bag X contains 4 red marbles and 2 blue marbles. Bag Y contains 3 red marbles and 4 blue marbles. Rahul also has a coin which is biased so that the probability of obtaining a head when it is thrown is 14 .
Rahul throws the coin.
• If he obtains a head, he chooses at random a marble from bag X. He notes the colour and replaces the marble in bag X. He then chooses at random a second marble from bag X.
• If he obtains a tail, he chooses at random a marble from bag Y. He notes the colour and discards the marble. He then chooses at random a second marble from bag Y.
(a) Find the probability that the two marbles that Rahul chooses are the same colour. [3]
(b) Find the probability that the two marbles that Rahul chooses are both from bag Y given that both marbles are blue. [3]
Mark scheme
4(a) 1 4 4 16 4 B1 2 clearly identified unsimplified probabilities from P(HRR),
P(HRR) = × × = , P(TRR), P(HBB) and P(TBB) correct.
4 6 6 144 36
3 3 2 18 3
P(TRR) = × × = ,
4 7 6 168 28
1 2 2 4 1
P(HBB) = × × = ,
4 6 6 144 36
3 4 3 36 6 3
P(TBB) = × × = , ,
4 7 6 168 28 14
4 3 1 6 M1 Sum of 4 correct scenarios, may be identified by the
+ + + unsimplified probability calculations.
36 28 36 28
29 A1 (0.460317… to at least 3SF).
= or 0.460
63
3
4(b) 3 4 3 M1 3 4 3 36
× × ,
3
oe, , 0.2142857 seen as numerator of a
P ( T BB ) 4 × 7 × 6
P ( T|BB ) =
4 7 6 168 14
=
P ( BB ) 1 + 6 fraction, accept unsimplified, FT their P(TBB) from 4(a)
36 28
M1 1 6
their + their FT from 4(a) or correct, 0.24206…,
3 36 28
3 61 seen as denominator of a fraction, accept unsimplified.
÷ or 14
14 252 61
252
54 A1 Accept 0.8852589… rounded to at least 3SF.
or 0.885 If one or both Ms not awarded, SC B1 for correct final
61
answer WWW.
3
The weights of the green apples sold by a shop are normally distributed with mean 90 grams and standard deviation 8 grams.
(a) Find the probability that a randomly chosen green apple weighs between 83 grams and 95 grams.
[4]
(b) The shop also sells red apples. 60% of the red apples sold by the shop weigh more than 80 grams.
160 red apples are chosen at random from the shop.
Use a suitable approximation to find the probability that fewer than 105 of the chosen red apples weigh more than 80 grams. [5]
Mark scheme
5(a) 83 − 90 95 − 90 M1 Using ± standardisation formula with 90, 8 and either 83 or
P(83 < X < 95) = P( <Z< )
8 8 95. Not σ 2 , not σ, no continuity correction.
= P (−0.875 < Z < 0.625) A1 Both ±0.875 OE and ±0.625 OE seen.
If M0 scored, SC B1 for both ±0.875 and ±0.625 seen
[ Φ ( 0.625) + Φ ( 0.875) −1 ] M1 Calculating the appropriate probability area, leading to their
final probability. Expect final answer > 0.5.
= 0.7340 + 0.8092 – 1
= 0.543 A1 0.5432, 0.543 ⩽ p < 0.5435.
Only dependent on the 2nd M mark.
4
5(b) [Mean =160 ×0.6 =]96 B1 96 and 38.4 seen, allow unsimplified.
[Var =160 ×0.6 × 0.4 =]38.4 May be seen in the standardisation formula.
8 15
, 6.19677 to at least 4 SF
5
implies correct variance
Withold mark if variance clearly identified as standard
deviation, condone N(96, 38.4 ) if standardisation formula
correct or variance/standard deviation correctly stated as
well.
104.5 − 96 M1 Substituting their 96 and their 38.4 into the ±standardising
P(X < 105) = P( Z < ) formula (any number for 104.5), condone σ2 or √σ.
38.4
M1 Use continuity correction 104.5 or 105.5 in their
[P( Z < 1.372) = Φ (1.372) ] standardisation formula.
±8.5 ±8.5
Note: or seen gains M2 BOD.
38.4 6.197
M1 Appropriate area Φ, from final process, must be a probability.
Expect final answer > 0.5.
= 0.915[0] A1 0.9149 ⩽ p ⩽ 0.915.
If one or more M marks not scored,
SC B1 for 0.9149 ⩽ p ⩽ 0.915.
The heights of the female students at Breven college are normally distributed:
• 90% of the female students have heights less than 182.7 cm.
• 40% of the female students have heights less than 162.5 cm.
(a) Find the mean and the standard deviation of the heights of the female students at Breven college.
[5]
Ten female students are chosen at random from those at Breven college.
(b) Find the probability that fewer than 8 of these 10 students have heights more than 162.5 cm. [3]
Mark scheme
6(a) 182.7 − μ B1 1.282 or – 1.282 seen, CAO (critical value).
= 1.282
σ
B1 −0.2535 < z ⩽ −0.253 or 0.253 ⩽ z < 0.2535 seen.
162.5 − μ
= −0.253
σ M1 One standardisation formula, not σ2, or √σ, with 182.7 or
162.5 substituted correctly equated to a z value (not 0.9, 0.1,
0.8159, 0.5398, 0.4, 0.6, 0.6554, 0.7257, …).
Solve, obtaining values for μ and σ M1 Either a single expression with one variable eliminated
formed or two expressions with both variables on the same
side seen with at least one variable value stated.
μ = 165.8, σ = 13.2 A1 Answers must be to at least 1 DP (context).
5
6(b) Method 1
[P(X < 8) = 1 – P(8, 9, 10) =] 1 – (10C8 ( 0.6 ) ( 0.4 ) + 10C9 x ( ) ( )
M1 One term 10C p x 1 − p 10− x . With 0 < p < 1, x ≠ 0 or 10.
8 2
( 0.6) ( 0.4) + ( 0.6) )
9 1 10
A1 Correct unsimplified expression. Allow 10 for 10C9.
Condone omission of last bracket only.
[= 1 – (0.12093 + 0.040311 + 0.0060466)] If both brackets omitted in unsimplified expression allow
recovery for final stated calculation of 1 – 0.1673 or final
answer WRT 0.8327.
= 0.833 B1 0.8327 < p ⩽ 0.833.
Method 2
( 0.4 ) +10C1 ( 0.6) ( 0.4) +10C2 ( 0.6) ( 0.4) +10C3 ( 0.6) ( 0.4) + x ( ) ( )
M1 One term 10C p x 1 − p 10− x . With 0 < p < 1, x ≠ 0 or 10.
10 1 9 2 8 3 7
C4 ( 0.6 ) ( 0.4 ) +10C5 ( 0.6 ) ( 0.4 ) + 10C6 ( 0.6 ) ( 0.4 ) + 10C7
10 4 6 5 5 6 4
A1 Correct unsimplified expression.
( 0.6) ( 0.4)
7 3
1.0486 ×10−4 + 1.5729 ×10−3 ++ 0.21499
= 0.833 B1 0.8327 < p ⩽ 0.833.
3
(a) How many different arrangements are there of the 9 letters in the word INTELLECT in which the two Ts are together? [2]
(b) How many different arrangements are there of the 9 letters in the word INTELLECT in which there is a T at each end and the two Es are not next to each other? [3]
Four letters are selected at random from the 9 letters in the word INTELLECT.
(c) Find the percentage of the possible selections which contain at least one E and exactly one T. [4]
Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
7(a) 8! M1 k!
k = 7 or 8, m = 1, 2.
2!2! 2! m !
= 10080 A1
7(b) Method 1 Number of ways with no restriction on Es – ways with Es together
7! 6! M1 7!
− – r, r integer > 1.
2!2! 2! 2!2!
[= 1260 – 360] M1 6!
s− , s integer > 360.
2!
= 900 A1
Method 2 T ^ ^ ^ ^ ^ T with Es inserted in gaps
5! 6 × 5 5! 6 M1 6×5
× or × C2 t× or t × 6C2 , t an integer > 1.
2! 2 2! 2
[= 60 × 15] M1 5!
× u , u an integer > 1.
2!
=900 A1
3
7(c) Method 1 – addition
T E _ _ = 2C1 × 2C1 × 5C2 = 40 B1 Either identified or correct unsimplified expression, either
alone or in an addition.
T E E _ = 2C1 × 2C2 × 5C1 = 10 B1 Either identified or correct unsimplified expression, either
alone or in an addition.
Probability
( 40 +10 ) M1 a
, a an integer < 126.
9 9
C4 C4
Denominator value must be seen as 9 C4 somewhere.
50 A1 39.68 ⩽ percentage ⩽ 39.7.
Percentage = 126 ×100 = 39.7%
Method 2 – subtraction (total arrangements with 1 T – number of arrangements with 1T 0 E)
T ^ ^ ^ = 2C1 × 7C3 = 70 B1 Either identified or correct unsimplified expression, either
alone or in a subtraction.
T * * * = 2C1 × 5C3 = 20 B1 Either identified or correct unsimplified expression, either
alone or in a subtraction.
Probability
( 70 − 20) M1 a
, a an integer < 126.
9 9
C4 C4
Denominator value must be seen as 9 C4 somewhere.
50 A1 39.68 ⩽ percentage ⩽ 39.7.
Percentage = 126 ×100 = 39.7%
4
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Mathematics 9709/51 for October/November 2024 is out of 50 marks, with 1 hour 15 minutes to answer it.
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