# CIE A level Math -Probability & Statistics 1: 5.4 Discrete random variables: calculate E(X) and Var(X) : Exam Style Questions Paper 5

### Question

A fair red spinner has edges numbered 1, 2, 2, 3. A fair blue spinner has edges numbered −3, −2, −1, −1.
Each spinner is spun once and the number on the edge on which each spinner lands is noted. The random variable X denotes the sum of the resulting two numbers.

(a) Draw up the probability distribution table for X.

(b)Given that E (X) = 0.25, find the value of Var (X).

(a) (b)\begin{aligned} & \frac{1}{16} \times-2^2+\frac{3}{16} \times-1^2\left(+\frac{5}{16} \times 0^2\right)+\frac{5}{16} \times 1^2+\frac{2}{16} \times 2^2-\left(\frac{1}{4}\right)^2 \\ & \frac{1 \times 4+3 \times 1+5 \times 0+5 \times 1+2 \times 4}{16}-0.25^2\end{aligned}

$\left[=\frac{5}{4}-\frac{1}{16}=\right] \frac{19}{16}, 1.1875$

### Question

A company produces small boxes of sweets that contain 5 jellies and 3 chocolates. Jemeel chooses
3 sweets at random from a box.
(a) Draw up the probability distribution table for the number of jellies that Jemeel chooses.               

The company also produces large boxes of sweets. For any large box, the probability that it contains
more jellies than chocolates is 0.64. 10 large boxes are chosen at random.
(b) Find the probability that no more than 7 of these boxes contain more jellies than chocolates.        

Ans

3 (a) (B1 for probability distribution table with correct outcome values)

$$P(0)=\frac{3}{8}\times \frac{2}{7}\times \frac{1}{6}=\frac{1}{56}$$

$$P(1)=\frac{5}{8}\times \frac{3}{7}\times \frac{2}{6}\times 3=\frac{15}{56}$$

$$P(2)=\frac{5}{8}\times \frac{4}{7}\times \frac{3}{6}\times 3=\frac{30}{56}$$

$$P(3)=\frac{5}{8}\times \frac{4}{7}\times \frac{3}{6}=\frac{10}{56}$$

(M1 for denominator 8×7×6)
Any one probability correct (with correct outcome)
All probabilities correct

3 (b) $$1-P(8,9,10)=1-[^{10}C_{8}0.64^{8}0.36^{2}+^{10}C_{9}0.64^{9}0.36^{1}+0.64^{10}]$$

1 – (0.164156 + 0.064852 + 0.11529)

0.759

### Question

A flower shop has 5 yellow roses, 3 red roses and 2 white roses. Martin chooses 3 roses at random. Draw up the probability distribution table for the number of white roses Martin chooses. 

Ans: ### Question

An ordinary fair die is thrown repeatedly until a 1 or a 6 is obtained.
(a) Find the probability that it takes at least 3 throws but no more than 5 throws to obtain a 1 or 6.
On another occasion, this die is thrown 3 times. The random variable X is the number of times that a 1 or a 6 is obtained.
(b) Draw up the probability distribution table for X.
(c) Find E(X).

Ans:

(a) $$(\frac{1}{3})(\frac{2}{3})^2+(\frac{1}{3})(\frac{2}{3})^3+(\frac{1}{3})(\frac{2}{3})^4$$
$$=\frac{4}{27}+\frac{8}{81}+\frac{16}{243}(=\frac{2432}{7776})$$
$$=\frac{76}{243}$$ or 0.313
(b)

 x 0 1 2 3 P(x) $$\frac{8}{27}$$ $$\frac{12}{27}$$ $$\frac{6}{27}$$ $$\frac{1}{27}$$

P(0) = $$(\frac{2}{3})^3$$
P(1) = $$(\frac{1}{3})(\frac{2}{3})^2 \times 3$$
P(2) = $$(\frac{2}{3})(\frac{1}{3})^2 \times 3$$
P(3) = $$(\frac{1}{3})^3$$
(c) $$E(X)=[0 \times \frac{8}{27}]+1 \times \frac{12}{27} + 2 \times \frac{6}{27} + 3 \times \frac{1}{27}$$
$$=[\frac{0}{27}]+\frac{12}{27}+\frac{12}{27}+\frac{3}{27}$$
= 1

### Question

Georgie has a red scarf, a blue scarf and a yellow scarf. Each day she wears exactly one of these
scarves. The probabilities for the three colours are 0.2, 0.45 and 0.35 respectively. When she wears a
red scarf, she always wears a hat. When she wears a blue scarf, she wears a hat with probability 0.4.
When she wears a yellow scarf, she wears a hat with probability 0.3.

(a) Find the probability that on a randomly chosen day Georgie wears a hat.

(b) Find the probability that on a randomly chosen day Georgie wears a yellow scarf given that she does not wear a hat.

Ans:

1. $$0.2[\times1]+0.45\times0.35\times0.3$$
0.485 or $$\frac{97}{200}$$
2. $$(Y|\bar{H})=\frac{P(Y\cap \bar{H})}{P(\bar{H})}=\frac{0.35\times 0.7}{1-their(a)}=\frac{0.245}{0.515}$$
0.476 or $$\frac{49}{103}$$

Two fair coins are thrown at the same time. The random variable X is the number of throws of the two coins required to obtain two tails at the same time.

### (a) Question

Find the probability that two tails are obtained for the first time on the 7th throw.

Ans:

$$\left ( \frac{3}{4} \right )^{6} \frac{1}{4}$$

$$0.0445,\frac{729}{16384}$$

### (b) Question

Find the probability that it takes more than 9 throws to obtain two tails for the first time.

Ans:

$$\left ( \frac{3}{4} \right )^{9}$$

$$0.0751,\frac{19683}{262144}$$

A fair spinner has edges numbered 0, 1, 2, 2. Another fair spinner has edges numbered −1, 0, 1. Each spinner is spun. The number on the edge on which a spinner comes to rest is noted. The random variable X is the sum of the numbers for the two spinners.

### (a) Question

Draw up the probability distribution table for X.

Ans: ### (b) Question

Find Var (x).

Ans:

$$E (X) = -\frac{1}{12}+\frac{4}{12}+\frac{6}{12}+\frac{6}{12} \left [ =\frac{15}{12} \right ]$$

$$Var (X) = \frac{1}{12}+0 + \frac{4}{12}+\frac{12}{12}+\frac{18}{12}-\left ( \frac{15}{12} \right )^{2}$$

$$\left [ \frac{35}{12} – \frac{25}{16} =\right ] \frac{65}{48}, 1.35$$

### Question

There are 400 students at a school in a certain country. Each student was asked whether they preferred
swimming, cycling or running and the results are given in the following table.

 Swimming Cycling Running Female 104 50 66 Male 31 57 92

A student is chosen at random.

(a) (i) Find the probability that the student prefers swimming.
(ii) Determine whether the events ‘the student is male’ and ‘the student prefers swimming’ are independent, justifying your answer.
On average at all the schools in this country 30% of the students do not like any sports.

(b) (i) 10 of the students do not like any sports.
Find the probability that at least 3 of these students do not like any sports.
(ii) 90 students from this country are now chosen at random.
Use an approximation to find the probability that fewer than 32 of them do not like any sports.

Ans:

1. (i) $$[\frac{104+31}{400}=]\frac{135}{400},\frac{27}{80},0.3375$$
(ii) Method 1
$$P(M)=\frac{180}{400},0.45 P(S)=\frac{135}{400},0.3375 P(M\cap S)=\frac{31}{400},0.0775$$
$$\frac{180}{400}\times \frac{135}{400}=\frac{243}{1600},0.151875≠\frac{31}{400}$$ so NOT independent
Method 2
$$P(M\cap S)=\frac{31}{400} P(S)=\frac{135}{400} P(M)=\frac{180}{400}$$
$$P(M|S)=\frac{\frac{31}{400}}{\frac{135}{400}}=\frac{31}{135},0.2296… ≠\frac{180}{400}$$ so NOT independent
2. (i) Method 1 [1-P(0, 1, 2)]
$$=1-(^{10}C_00.3^00.7^{10}+^{10}C_10.3^10.7^9+^10C_20.3^20.7^8)$$
=1-(0.028248+0.131061+0.233474)
Method 2 [P(3, 4, 5,6,7,8,9,10)=]
$$^{10}C_30.3^30.7^7+^{10}C_40.3^40.7^6+^{10}C_50.3^50.7^5+^{10}C_60.3^60.7^4+^{10}C_70.3^70.7^3+^{10}C_80.3^80.7^2+^{10}C_90.3^90.7^1+^{10}C_{10}0.3^{10}0.7^0$$
=0.617
[p=0.3]
Mean $$= 0.3 \times 90=27;$$
variance $$=0.3 \times 90 \times 0.7 = 18.9$$
$$P(X<32)=P(z<\frac{31.5-27}{\sqrt{18.9}})$$
=Φ(1.035)
=0.850

### Question

The random variable X takes each of the values 1, 2, 3, 4 with probability ¼. 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.                                                                                                   

(b) Find the probability that Y = 2 given that Y is even.                                                                                             

4 (a) 4 (b)  $$P(2|even)=\frac{\frac{5}{16}}{\frac{6}{16}}$$
$$\frac{5}{6}$$ or 0·833