CAIE A-Level Mathematics AS 5 Probability and Statistics 1 Questions

Practise Probability & Statistics 1 through data representation, counting, probability, discrete random variables and normal-distribution models using exact exam methods.

Covered syllabus points

5.1 Representation of data 5.2 Permutations and combinations 5.3 Probability 5.4 Discrete random variables 5.5 The normal distribution

[Maximum number: 10]

The heights, in cm, of the 11 basketball players in each of two clubs, the Amazons and the Giants, are shown below.

Table for Question 1 — CAIE A-Level Mathematics AS

(a)

State an advantage of using a stem-and-leaf diagram compared to a box-and-whisker plot to illustrate this information.

[ 1 ]

(b)

Represent the data by drawing a back-to-back stem-and-leaf diagram with Amazons on the left-hand side of the diagram.

[ 4 ]

(c)

Find the interquartile range of the heights of the players in the Amazons.

Four new players join the Amazons. The mean height of the 15 players in the Amazons is now 191.2 cm . The heights of three of the new players are $180 \mathrm{~cm}, 185 \mathrm{~cm}$ and 190 cm .

[ 2 ]

$\mathrm{UQ}=202\text{ cm},\qquad \mathrm{LQ}=182\text{ cm}, \qquad \mathrm{IQR}=202-182=20\text{ cm}.$

(d)

Find the height of the fourth new player.

[ 3 ]

$\Sigma_{11}=2132,\qquad \Sigma_{15}=191.2(15)=2868.$
Thus,
$2868=2132+180+185+190+h, \qquad h=181\text{ cm}.$

[Maximum number: 10]

A new village social club has 10 members of whom 6 are men and 4 are women. The club committee will consist of 5 members.

(a)

In how many ways can the committee of 5 members be chosen if it must include at least 2 men and at least 1 woman?

[ 4 ]

The valid committee compositions are:
$\binom64\binom41+\binom63\binom42+\binom62\binom43 =60+120+60=240.$

(b)

The 10 members of the club stand in a line for a photograph.

How many different arrangements are there of the 10 members if all the men stand together and all the women stand together?

[ 2 ]

Arrange the six men as a block and the four women as a block. The two blocks can be ordered in either way:
$6!\times4!\times2=34560.$

(c)

For a second photograph, the members stand in two rows, with 6 on the back row and 4 on the front row. Olly and his sister Petra are two of the members of the club. How many different arrangements are there of the 10 members in which Olly and Petra stand next to each other on the front row?

[ 4 ]

Olly and Petra must be in the front row. Arrange them in 2! orders, choose and arrange two of the remaining eight people in the other front-row positions, and arrange the remaining six in the back row:
$\binom82\times3!\times2!\times6!=241920.$

[Maximum number: 8]

(a)

How many different arrangements are there of the 9 letters in the word RECORDERS?

[ 1 ]

The repeated letters are the three R's and the two E's. Hence the number of arrangements is
$\frac{9!}{2!\,3!}=30240.$

(b)

How many different arrangements are there of the 9 letters in the word RECORDERS in which there is an E at the beginning, an E at the end and the three Rs are not all together?

[ 3 ]

With an E at each end, the number of arrangements is 7!/3!. If the three R's are together, the number is 5!. Therefore,
$\frac{7!}{3!}-5!=840-120=720.$

(c)

The 9 letters of the word RECORDERS are divided at random into two groups: a group of 5 letters and a group of 4 letters.

Find the probability that the three Rs are in the same group.

[ 4 ]

The favourable cases are:

- Group of 5 contains the three R's and two E's: 1 way.
- Group of 5 contains the three R's and one E: $\binom21\binom41=8$ ways.
- Group of 5 contains the three R's and no E: $\binom42=6$ ways.
- Group of 4 contains the three R's and one E: $\binom21=2$ ways.
- Group of 4 contains the three R's and no E: $\binom41=4$ ways.

Thus the total number of favourable splits is 21. The total number of splits is $\binom95=126$ , so
$P(\text{three }R\text{'s in the same group})=\frac{21}{126}=\frac16=0.167.$

[Maximum number: 8]

The numbers on the faces of a fair six-sided dice are 1,2,2,3,3,3. The random variable X is the total score when the dice is rolled twice.

(a)

Draw up the probability distribution table for X.

[ 3 ]

The probability distribution is

x	2	3	4	5	6
P(X=x)	$\frac{1}{36}$	$\frac{4}{36}=\frac19$	$\frac{10}{36}=\frac5{18}$	$\frac{12}{36}=\frac13$	$\frac{9}{36}=\frac14$

(b)

Find the value of $\operatorname{Var}(X)$ .

[ 3 ]

The mean is
$E(X)=\frac{1(2)+4(3)+10(4)+12(5)+9(6)}{36}=\frac{14}{3}.$
Therefore,
$\begin{aligned} \operatorname{Var}(X) &=\frac{1(2^2)+4(3^2)+10(4^2)+12(5^2)+9(6^2)}{36}-\left(\frac{14}{3}\right)^2\\ &=\frac{10}{9}. \end{aligned}$

(c)

Find the probability that X is even given that X>3.

[ 2 ]

Given that X>3, the possible values are 4,5,6. Thus,
$P(X\text{ is even}\mid X>3) =\frac{\frac{10}{36}+\frac{9}{36}}{\frac{31}{36}} =\frac{19}{31}=0.613.$

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Covered syllabus points

Question 1

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Question (b)

Question (c)

Question (d)

Question 2

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Question 3

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Question 4

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