Additional Problems - Chapter 6 Random Variable and Mathematical Expectation 12th Maths Guide Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Problems
One Mark Questions
Question 1.
If a fair coin is tossed three times the probability function $p(x)$ of the number of heads $\mathrm{x}$ is
(a)

(b)

(c)

(d) None of these
Answer:
(b)

Hint:
The sample space is HHH, HHT, HTH, HTT, THH, THT, TTH and TTT Number of heads is $0,1,2,3$ with probability $\frac{1}{8}, \frac{3}{8}, \frac{3}{8}$ and $\frac{1}{8}$
Question 2.
If a discrete random variable has the probability mass function as

then the value of $\mathrm{K}$ is
(a) $\frac{1}{11}$
(b) $\frac{1}{12}$
(c) $\frac{2}{13}$
(d) $\frac{4}{9}$
Answer:
(b) $\frac{1}{12}$
Hint:
$
\begin{aligned}
& \mathrm{k}+3 \mathrm{k}+6 \mathrm{k}+2 \mathrm{k}=1 \\
& \Rightarrow 12 \mathrm{k}=1 \\
& \Rightarrow \mathrm{k}=\frac{1}{12}
\end{aligned}
$
Question 3.
If the probability density function of $X$ is $f(x)=C x(2-x)$, and $0 (a) $\frac{4}{3}$
(b) $\frac{6}{5}$
(c) $\frac{3}{4}$
(d) $\frac{3}{5}$
Answer:
(c) $\frac{3}{4}$
Hint:
$
\begin{aligned}
& \int_0^2 f(x) d x=1 \\
& \Rightarrow \int_0^2 C\left(2 x-x^2\right) d x \Rightarrow C\left(x^2-\frac{x^3}{3}\right)_0^2=1 \\
& \Rightarrow C\left[4-\frac{8}{3}\right]=1 \Rightarrow \frac{4}{3} C=1 \text { (or) } C=\frac{3}{4}
\end{aligned}
$
Question 4.
The random variables $\mathrm{X}$ and $\mathrm{Y}$ are independent if

(a) $\mathrm{E}(\mathrm{XY})=1$
(b) $\mathrm{E}(\mathrm{XY})=0$
(c) $\mathrm{E}(\mathrm{XY})=\mathrm{E}(\mathrm{X}) \mathrm{E}(\mathrm{Y})$
(d) $\mathrm{E}(\mathrm{X}+\mathrm{Y})=\mathrm{E}(\mathrm{X})+\mathrm{E}(\mathrm{Y})$
Answer:
(c) $\mathrm{E}(\mathrm{XY})=\mathrm{E}(\mathrm{X}) \mathrm{E}(\mathrm{Y})$
Question 5.
If a random variable $\mathrm{X}$ has the following distribution

then the expected value of $\mathrm{X}$ is
(a) $\frac{3}{2}$
(b) $\frac{1}{6}$
(c) $\frac{1}{2}$
(d) $\frac{1}{3}$
Answer:
(b) $\frac{1}{6}$
Hint:
$
\mathrm{E}(\mathrm{X})=\frac{-1}{3}-\frac{2}{6}+\frac{1}{6}+\frac{2}{3}=\frac{1}{3}-\frac{1}{6}=\frac{1}{6}
$
Question 6.
$\operatorname{Var}(4 \mathrm{X}+7)=$
(a) 7
(b) 16 Var (X)
(c) 11
(d) None of these
Answer:
(b) 16 Var (X)
Hint:
$\operatorname{Var}(4 \mathrm{X}+7)=(42) \operatorname{Var}(\mathrm{X})$
Question 7.
Match the following:

Answer:
(a) - (iii)
(b) - (iv)
(c) - (v)
(d) - (ii)
(e) - (i)
Question 8.

If $\mathrm{E}(\mathrm{X})=2$ and $\mathrm{E}(\mathrm{Z})=4$, then $\mathrm{E}(\mathrm{Z}-\mathrm{X})$ is
(a) 2
(b) 6
(c) 0
(d) $-2$
Answer:
(a) 2
Hint:
$
\mathrm{E}(\mathrm{Z}-\mathrm{X})=\mathrm{E}(\mathrm{Z})-\mathrm{E}(\mathrm{X})=4-2=2
$
Question 9.
Fill in the blanks:
1. The distribution function $F(X)$ is equal to __________
2. Two types of random variables are _________ and _____
3. Probability mass function is also called _____
4. Cumulative distribution function is also called ___________
5. Probability density function is also called __________
6. $\mathrm{d} \mathrm{F}(\mathrm{x})$ is known as _____________ of $\mathrm{X}$.
7. $\mathrm{E}(\mathrm{X})$ is denoted by ___________
8. Variance is a measure of or of $X$.
9. Standard deviation is defined as _________ and _____________
10. Mean is the __________ of a density. Variance is the ______________ of a density.
Answers:
1. $\mathrm{P}(\mathrm{X} \leq \mathrm{x})$
2. discrete and continuous
3. discrete probability function
4. distribution function
5. continuous probability function, integrating the density function
6. probability differential
7. $\mu_{\mathrm{x}}$
8. the spread, dispersion of the density

9. $\sqrt{\operatorname{Var}[\mathrm{X}]}$
10. center of gravity, the moment of inertia.
2 Mark Questions
Question 1.
Verify whether the following function is a probability mass function or not. Hence find c.d.f.
$
p(x)=\left\{\begin{array}{l}
\frac{1}{3}, \text { for } x=1 \\
\frac{2}{3}, \text { for } x=2 \\
0, \text { otherwise }
\end{array}\right.
$
Solution:
$
\Sigma \mathrm{p}_{\mathrm{i}}=\frac{1}{3}+\frac{2}{3}=1 \text { and } \mathrm{p}_{\mathrm{i}}>0 \text {. }
$
So the given function is a p.m.f. c.d.f is given by
$
\begin{aligned}
& \mathrm{F}(\mathrm{x})=\mathrm{P}(\mathrm{X} \leq \mathrm{x}) \\
& \mathrm{F}(1)=\mathrm{P}(\mathrm{X} \leq 1)=\mathrm{P}(\mathrm{X}=1)=\frac{1}{3} \\
& \mathrm{~F}(2)=\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)=\frac{1}{3}+\frac{2}{3}=1 \\
&
\end{aligned}
$
Question 2.
Consider the following probability distribution of $\mathrm{X}$.

Is $\mathrm{p}\left(\mathrm{x}_{\mathrm{i}}\right)$ a p.m.f?
Solution:
$\mathrm{p}\left(\mathrm{x}_{\mathrm{i}}\right)>0$ for all $\mathrm{i}$
$
\begin{gathered}
\sum p\left(x_i\right)=\frac{1}{36}+\frac{2}{36}+\frac{3}{36}+\frac{4}{36}+\frac{5}{36}+\frac{6}{36}+\frac{5}{36}+\frac{4}{36}+\frac{3}{36}+\frac{2}{36}+\frac{1}{36} \\
=\frac{36}{36}=1
\end{gathered}
$
Hence $p\left(x_1\right)$ is a p.m.f.

Question 3.
The probability that a man fishing at a particular place will catch $1,2,3$ and 4 fish are $0.4,0.3,0.2$ and $0.1$. What is the expected number of fish caught?
Solution:
Let $\mathrm{X}$ denote the no.of fish caught by the man.

$
\begin{aligned}
& E(X)=1(0.4)+2(0.3)+3(0.2)+4(0.1) \\
& =0.4+0.6+0.6+0.4 \\
& =2
\end{aligned}
$
Question 4.
A random variable $\mathrm{X}$ has the following probability function.

Find the value of $K$.
Solution:
We know that $\Sigma \mathrm{p}\left(\mathrm{x}_{\mathrm{i}}\right)=1$
$
\begin{aligned}
& \Rightarrow 0.1+\mathrm{K}+0.2+2 \mathrm{~K}+0.3+\mathrm{K}=1 \\
& \Rightarrow 4 \mathrm{~K}+0.6=1 \\
& \Rightarrow 4 \mathrm{~K}=0.4 \\
& \Rightarrow \mathrm{K}=0.1
\end{aligned}
$
Question 5.
A person receives a sum of rupees equal to the square of the number that appears on the face when a die is tossed. How much money can he expect to receive?
Solution:
Let the random variable $\mathrm{X}$ denote the square of the number that can appear on the face of a die. Then the distribution is given by

$
\begin{aligned}
\mathrm{E}(\mathrm{X}) & =1^2\left(\frac{1}{6}\right)+2^2\left(\frac{1}{6}\right)+3^2\left(\frac{1}{6}\right) \\
& =\frac{1}{6}[1+4+9+16+25+36] \\
& =\frac{1}{6}[91]=\frac{91}{6}
\end{aligned}
$
He can expect to receive Rs. $\frac{91}{6}$

3 and 5 Marks Questions
Question 1.
For the following distribution of $\mathrm{X}$.

(i) $\mathrm{P}(\mathrm{X} \leq 1)$
(ii) $\mathrm{P}(\mathrm{X} \leq 2)$
(iii) $\mathrm{P}(0<\mathrm{X}<2)$
Solution:
(i) $\mathrm{P}(\mathrm{X} \leq 1)=\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=0)$
$=\frac{1}{2}+\frac{1}{6}=\frac{4}{6}=\frac{2}{3}$
(ii) $\mathrm{P}(\mathrm{X} \leq 2)=\mathrm{P}(\mathrm{X}=2)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=0)$
$=\frac{3}{10}+\frac{1}{2}+\frac{1}{6}=\frac{29}{30}$
(iii) $\mathrm{P}(0<\mathrm{X}<2)=\mathrm{P}(\mathrm{X}=1)=\frac{1}{2}$
Question 2.
Given that p.d.f of a random variable $X$ as follows $f(x)= \begin{cases}k x(1-x), \text { for } 0 Find $\mathrm{K}$ and c.d.f.
Solution:
We know that if $f(x)$ is a p.d.f, then $\int_{-\infty}^{\infty} f(x) d x=1$
$
\begin{gathered}
\therefore \int_0^1 k x(1-x) d x=1 \Rightarrow k \int_0^1\left(x-x^2\right) d x=1 \\
\Rightarrow k\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1=1 \Rightarrow k\left(\frac{1}{2}-\frac{1}{3}\right)=1 \Rightarrow \frac{k}{6}=1 \\
\text { or } k=6
\end{gathered}
$
Hence the given p.d.f becomes,
$
f(x)=\left\{\begin{array}{lc}
6 x(1-x) & \text { for } 0 0, & \text { otherwise }
\end{array}\right.
$

To find c.d.f $\mathrm{F}(x)$,
$
\begin{aligned}
& \begin{aligned}
\mathrm{F}(x)=0 \quad \text { for } x & \leq 0 \\
\mathrm{~F}(x)=\mathrm{P}(\mathrm{X} \leq x) & =\int_{-\infty}^x f(x) d x \\
& =\int_0^x 6 x(1-x) d x=\int_0^x\left(6 x-6 x^2\right) d x \\
& =\left[\frac{6 x^2}{2}-\frac{6 x^3}{3}\right]_0^x=3 x^2-2 x^3
\end{aligned} \\
& \mathrm{~F}(x)=1 \text { for } x \geq 1
\end{aligned}
$
$\therefore$ The c.d.f of $\mathrm{X}$ is as follows:
$
\begin{aligned}
F(x) & =0 \text { for } x \leq 0 \\
& =3 x^2-2 x^3 \text { for } 0 & =1 \quad \text { for } x \geq 1
\end{aligned}
$
Question 3.
Suppose that the life in hours of a certain part of radio tube is an r.v $\mathrm{X}$ with p.d.f given by
$
f(x)=\left\{\begin{array}{l}
\frac{100}{x^2}, x \geq 100 \\
0, \text { otherwise }
\end{array}\right.
$
(i) What is the probability that all of three Such tubes in a given radio set will have to be replaced in the first 150 hours?
(ii) What is the probability that none of the three tubes will be replaced?
Solution:
(i) A tube in the radio set will have to be replaced during the first 150 hours if its life is less than 150 hours. Hence the required probability is
$
\begin{aligned}
\mathrm{P}(\mathrm{X} \leq 150) & =\int_{100}^{150} f(x) d x=\int_{100}^{150} \frac{100}{x^2} d x \\
& =\left(\frac{-100}{x}\right)_{100}^{150}=\frac{-100}{150}+1=\frac{1}{3}
\end{aligned}
$

The probability that all three of the original tubes will have to be replaced during the first 150 hours is $\left(\frac{1}{3}\right)^3=\frac{1}{27}$
(ii) The probability that a tube is not replaced is given by $\mathrm{P}(\mathrm{X}>150)$
$=1-\mathrm{P}(\mathrm{X} \leq 150)$
$=1-\frac{1}{3}$
$=\frac{2}{3}$
Hence the probability that none of the three tubes will be replaced during the 150 hours of operation is $\left(\frac{2}{3}\right)^3=\frac{8}{27}$
Question 4.
Let $\mathrm{X}$ be a continuous random variable with p.d.f:

(i) Find 'a'
(ii) compute $\mathrm{P}(\mathrm{X} \leq 1.5)$
Solution:
Since $f(x)$ is a p.d.f $\int_{-\infty}^{\infty} f(x) d x=1$
$\Rightarrow \int_0^1(a x) d x+\int_1^2 a d x+\int_2^3(-a x+3 a) d x=1$
$\left(\frac{a x^2}{2}\right)_0^1+(a x)_1^2+\left[\frac{-a x^2}{2}+3 a x\right]_2^3=1$
$\frac{a}{2}+2 a-a-\frac{9}{2} a+9 a+2 a-6 a=1$
$\frac{a}{2}-\frac{9 a}{2}+6 a=1$
$2 a=1$
$\Rightarrow \quad a=\frac{1}{2}$

$
\begin{aligned}
\mathrm{P}(\mathrm{X} \leq 1.5) & =\int_0^1 a x d x+\int_1^{1.5} a d x \\
& =\left(\frac{a x^2}{2}\right)_0^1+(a x)_1^{1.5} \\
& =\frac{a}{2}+1.5 a-a=\frac{a}{2}+\frac{a}{2}=a \\
& =\frac{1}{2}\left(\text { since } a=\frac{1}{2}\right)
\end{aligned}
$
Question 5.
A random variable $\mathrm{X}$ has the probability function as follows:

Find $\mathrm{E}(3 \mathrm{X}+1), \mathrm{E}\left(\mathrm{X}^2\right)$ and $\operatorname{Var}(\mathrm{X})$.
Solution:
$
\begin{aligned}
& \mathrm{E}(\mathrm{X})=(-1)(0.2)+0(0.3)+1(0.5)=-0.2+0.5=0.3 \\
& \text { So } \mathrm{E}(3 \mathrm{X}+1)=3 \mathrm{E}(\mathrm{X})+1=3(0.3)+1=1.9 \\
& \mathrm{E}\left(\mathrm{X}^2\right)=(-1)^2(0.2)+0^2(0.3)+1^2(0.5)=0.2+0.5=0.7 \\
& \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2=0.7-(0.3)^2=0.61
\end{aligned}
$
Question 6.
A player tossed two coins. If two heads show he wins Rs.4. If one head shows he wins Rs.2, but if two tails show he must pay Rs. 3 as a penalty. Calculate the expected value of " the sum won by him.
Solution:
Let $\mathrm{X}$ be the discrete random variable denoting the sum won by the player. We know that,
Probability of getting 2 heads $=\frac{1}{4}$
Probability of getting 1 head is $\frac{1}{2}$
Probability of getting 2 tail is $\frac{1}{4}$
So the player wins Rs. 4 with probability $\frac{1}{4}$, he wins Rs. 2 with probability $\frac{1}{2}$ and loses Rs. 3 with probability $\frac{1}{4}$

$
\begin{aligned}
& \mathrm{E}(\mathrm{X})=4\left(\frac{1}{4}\right)+2\left(\frac{1}{2}\right)-3\left(\frac{1}{4}\right) \\
& =1+1-\frac{3}{4} \\
& =\frac{5}{4}
\end{aligned}
$
Thus the expected value of the sum won by him is Rs.1.25.