| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Construct probability distribution from scenario |
| Difficulty | Moderate -0.8 This is a straightforward S1 question requiring basic counting principles and systematic enumeration. Part (i) is routine combinatorics (choosing 2 from 3 even, 1 from 2 odd), and part (ii) involves listing all C(5,3)=10 combinations and identifying the minimum—mechanical work with no conceptual challenge or novel insight required. |
| Spec | 2.04a Discrete probability distributions5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks |
|---|---|
| (ii) | 3 2 2 3 |
| Answer | Marks |
|---|---|
| P(S =s) 6/10 3/10 1/10 | M1 |
| Answer | Marks |
|---|---|
| B1 5 | 5×4×3 seen in denom |
| Answer | Marks |
|---|---|
| (b) | N*B |
| Answer | Marks |
|---|---|
| Other in C8 = 495 | B1 |
| Answer | Marks |
|---|---|
| B1 | 5! Seen in num oe or alone mult by k [ 1 |
| Answer | Marks | Guidance |
|---|---|---|
| s | 1 | 2 |
| P(S =s) | 6/10 | 3/10 |
| Page 6 | Mark Scheme | Syllabus |
| Cambridge International AS/A Level – May/June 2015 | 9709 | 62 |
| Answer | Marks |
|---|---|
| (ii) | 30−35.2 |
| Answer | Marks |
|---|---|
| t = 37.0 | M1 |
| Answer | Marks |
|---|---|
| A1 3 | Standardising no sq rt no cc no sq |
| Answer | Marks |
|---|---|
| (b) | 7−µ =−0.8σ |
| Answer | Marks |
|---|---|
| µ =8.94 σ =2.42 | B1 |
| Answer | Marks |
|---|---|
| A1 5 | ± 0.8 seen |
Question 5:
--- 5 (i)
(ii) ---
5 (i)
(ii) | 3 2 2 3
3
P(2Es 1O) = × × × C2 = (0.6)
5 4 3 5
OR
3C ×2C 6
2 1
P(2Es 1O) = =
5C 10
3
= 0.6
OR
241, 247, 261, 267, 461, 467 = 6 options
124 126 127 146 147 167 246 247 267 467
Prob = 6/10
124 126 127 146 147 167
246 247 267 467
s 1 2 4
P(S =s) 6/10 3/10 1/10 | M1
M1
A1 3
M1
M1
A1
M1
M1
A1
M1
A1
B1
B1
B1 5 | 5×4×3 seen in denom
3
Mult a prob by C2 oe
Correct answer
3 Cx or y C2 or 2 C1 oe seen mult by k [ 1 in
num
5
C3 seen in denom
Correct answer
List at least 3 of 241, 247, 261, 267, 461, 467
5
C3 or list to get all 10 options in denom
see below
Correct answer
Attempt at listing with at least 7 correct
All correct and no others or all 60
1, 2, 4 only seen in top row
Any two correct
All correct
6 (a) (i)
(ii)
(b) | N*****B
5!
Number of ways =
3!
= 20
B(AAA)NNS
5!
5
Number of ways = or P3
2!
= 60
14
C9 total options = 2002
12
T and M both in C7 = 792
Ans 2002 – 792 = 1210
OR
12
Neither in C9 = 220
12
One in C8 = 495
12
Other in C8 = 495 | B1
B1
B1 3
M1
M1
A1 3
M1
B1
A1 3
M1
B1 | 5! Seen in num oe or alone mult by k [ 1
3! Seen in denom can be mult by k [ 1
Correct final answer
5! seen as a num can be mult by k [ 1
Dividing by 2!
Correct final answer
14 14
C9 or P9 in subtraction attempt
12
C7 (792) seen
Correct final answer
Summing 2 or 3 options at least 1 correct
12 12 12
condone P9 + P8 + P8 here only
Second correct option seen accept another
495 or if M1 not awarded, any correct
option
s | 1 | 2 | 4
P(S =s) | 6/10 | 3/10 | 1/10
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International AS/A Level – May/June 2015 | 9709 | 62
7 (a) (i)
(ii) | 30−35.2
prob = p z<
4.7
= P( z < –1.106)
= 1 – 0.8655 = 0.1345
0.1345×52 = 6.99
Φ(t) = 0.648 z = 0.380
t −35.2
0.380 =
4.7
t = 37.0 | M1
M1
A1
A1 4
B1
M1
A1 3 | Standardising no sq rt no cc no sq
1−Φ
Correct ans rounding to 0.13
Correct final answer accept 6 or 7 if 6.99
not seen but previous prob 0,1345 correct
0.648 seen
standardising allow cc, sq rt,sq, need use of
tables not 0.148, 0.648, 0.352, 0.852
correct answer rounding to 37.0
(b) | 7−µ =−0.8σ
so 7−µ =−0.8σ
σ
10−µ
=0.44 so 10−µ =0.44σ
σ
µ =8.94 σ =2.42 | B1
B1
M1
M1
A1 5 | ± 0.8 seen
± 0.44 seen
An eqn with z-value, µ and σ no sq rt no cc
no sq
Sensible attempt to eliminate µ or σ by
subst or subtraction, need at least one value
Correct answersA box contains 5 discs, numbered 1, 2, 4, 6, 7. William takes 3 discs at random, without replacement, and notes the numbers on the discs.
\begin{enumerate}[label=(\roman*)]
\item Find the probability that the numbers on the 3 discs are two even numbers and one odd number. [3]
\end{enumerate}
The smallest of the numbers on the 3 discs taken is denoted by the random variable $S$.
\begin{enumerate}[label=(\roman*), resume]
\item By listing all possible selections (126, 246 and so on) draw up the probability distribution table for $S$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE S1 2015 Q5 [8]}}