Exercise 11.6-Additional Problems - Chapter 11 - Probability Distributions - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Questions Solved
Choose the Correct or the most suitable answer from the given four alternatives:
Question 1.
If $f(x)=\left\{\begin{array}{cc}k x^2, & 0 (a) $\frac{1}{3}$
(b) $\frac{1}{6}$
(c) $\frac{1}{9}$
(d) $\frac{1}{12}$
Solution:
(c) $\frac{1}{9}$
Hint:
Given that $\mathrm{f}(\mathrm{x})$ is a probability density function
$
\begin{aligned}
& \therefore \int_{-\infty}^{\infty} f(x) d x=1 \\
& \int_{-\infty}^0 f(x) d x+\int_0^3 f(x) d x+\int_3^{\infty} f(x) d x=1 \\
& 0+\int_0^3 k x^2 d x+0=1 \Rightarrow k \cdot\left[\frac{x^3}{3}\right]_0^3=1 \\
& \frac{k}{3}\left[3^3-0^3\right]=1 \\
& \frac{27 k}{3}=1 \Rightarrow k=\frac{3}{27}=\frac{1}{9}
\end{aligned}
$
Question 2.
If $f(x)=\frac{\mathrm{A}}{\pi} \frac{1}{16+x^2},-\infty (a) 16
(b) 8
(c) 4
(d) 1
Solution:
(c) 4
Hint:
Given that $f(x)=\frac{\mathrm{A}}{\pi}$ is a probabiltiy density function
$
\begin{aligned}
& \therefore \int_{-\infty}^{\infty} f(x) d x=1 \\
& \int_{-\infty}^{\infty} \frac{\mathrm{A}}{\pi} \cdot \frac{1}{16+x^2} \cdot d x=1 \Rightarrow \frac{\mathrm{A}}{\pi} \int_{-\infty}^{\infty} \frac{d x}{4^2+x^2}=1
\end{aligned}
$
$\frac{2 \mathrm{~A}}{\pi} \int_0^{\infty} \frac{d x}{4^2+x^2}=1$, Since $\frac{1}{4^2+x^2}$ is an even function
$
\begin{aligned}
& \frac{2 \mathrm{~A}}{\pi} \cdot\left[\frac{1}{4} \cdot \tan ^{-1}\left(\frac{x}{4}\right)\right]_0^{\infty}=1 \Rightarrow \frac{\mathrm{A}}{2 \pi}\left[\tan ^{-1}(\infty)-\tan ^{-1} 0\right]=1 \\
& \frac{\mathrm{A}}{2 \pi}\left[\frac{\pi}{2}-0\right]=1 \Rightarrow \frac{\mathrm{A}}{4}=1 \Rightarrow \mathrm{A}=4
\end{aligned}
$

Question 3.
A random variable $\mathrm{X}$ has the following probability mass function as follows:

Then the value of $\lambda$ is .......
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2
Hint:
$
\begin{aligned}
& \Sigma \mathrm{P}_i=1, \\
& \frac{\lambda}{6}+\frac{\lambda}{4}+\frac{\lambda}{12}=1 \Rightarrow \frac{2 \lambda+3 \lambda+\lambda}{12}=1 \Rightarrow \frac{6 \lambda}{12}=1 \Rightarrow \frac{\lambda}{2}=1 \Rightarrow \lambda=2
\end{aligned}
$

Question 4.
$\mathrm{X}$ is a discrete random variable which takes the values of $0,1,2$ and $\mathrm{P}(\mathrm{X}=0)=\frac{144}{169}$, $P(X=1)=\frac{1}{169}$, then the value of $P(X=2)$ is
(a) $\frac{145}{169}$
(b) $\frac{24}{169}$
(c) $\frac{2}{169}$
(d) $\frac{143}{169}$
Solution:
(b) $\frac{24}{169}$
$
\begin{gathered}
\Sigma P_i=1 \\
P(X=0)+P(X=1)+P(X=2)=1 \\
\frac{144}{169}+\frac{1}{169}+P(X=2)=1 \\
\Rightarrow P(X=2)=1-\frac{145}{169}=\frac{169-145}{169}=\frac{24}{169}
\end{gathered}
$
Question 5.
Given $\mathrm{E}(\mathrm{x}+\mathrm{c})=8$ and $\mathrm{E}(\mathrm{x}-\mathrm{c})=12$, then the value of $\mathrm{c}$ is ......
(a) $-2$
(b) 4
(c) $-4$
(d) 2
Solution:
(a) $-2$
Hint:
$
\begin{aligned}
& \mathrm{E}(\mathrm{X}+\mathrm{c})=8 \\
& \mathrm{E}(\mathrm{X})+\mathrm{E}(\mathrm{c})=8 \\
& \mathrm{E}(\mathrm{X})+\mathrm{c}=8 \ldots . \\
& \mathrm{E}(\mathrm{X}-\mathrm{c})=12 \\
& \mathrm{E}(\mathrm{X})-\mathrm{E}(\mathrm{c})=12 \Rightarrow \mathrm{E}(\mathrm{X})-\mathrm{c}=12 \\
& (1)-(2) \Rightarrow 2 \mathrm{c}=-4=\mathrm{c}=-2
\end{aligned}
$

Question 6.
$\mathrm{X}$ is random variable taking the values 3,4 and 12 with probabilities $\frac{1}{3}, \frac{1}{4}$ and $\frac{5}{12}$. Then $\mathrm{E}(\mathrm{X})$ is $\ldots \ldots$...
(a) 5
(b) 7
(c) 6
(d) 3
Solution:
(b) 7
Hint:

$\mathrm{E}(\mathrm{X})=\Sigma x_i p_i$
$
E(X)=3\left(\frac{1}{3}\right)+4\left(\frac{1}{4}\right)+12\left(\frac{5}{12}\right)=1+1+5=7
$

Question 7.
Variance of the random variable $\mathrm{X}$ is 4 . Its mean is 2 . Then $\mathrm{E}\left(\mathrm{X}^2\right)$ is
(a) 2
(b) 4
(c) 6
(d) 8
Solution:
(d) 8
Hint:
Given, $\operatorname{Var}(\mathrm{X})=4$
$
\begin{aligned}
& \mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2=4 \\
& \mathrm{E}\left(\mathrm{X}^2\right)-2^2=4 \Rightarrow \mathrm{E}\left(\mathrm{X}^2\right)=4+4=8
\end{aligned}
$
Question 8.
In 5 throws of a die, getting 1 or 2 is a success. The mean number of success is .....
(a) $\frac{5}{3}$
(b) $\frac{3}{5}$
(c) $\frac{5}{9}$
(d) $\frac{9}{5}$

Solution:

(a) $\frac{5}{3}$
Hint:
Given, $\mathrm{n}=5$
Given, $n=5$
Getting 1 or 2 is a success
$
\begin{aligned}
p & =\frac{2}{6}=\frac{1}{3} \\
\text { mean } & =n p=5 \times \frac{1}{3}=\frac{5}{3}
\end{aligned}
$
Question 9.
The mean of a binomial distribution is 5 and its standard deviation is 2 . Then the value of $n$ and $p$ are
(a) $\left(\frac{4}{5}, 25\right)$
(b) $\left(25, \frac{4}{5}\right)$
$\left(\frac{1}{5}, 25\right)$
$\left(25, \frac{1}{5}\right)$
Solution:

(d) $\left(25, \frac{1}{5}\right)$
Hint:
$
\begin{aligned}
\text { Given, mean }=n p & =5 \\
\text { standard deviation }=\sqrt{n p q} & =2 \\
n p q & =4 \\
\frac{n p q}{n p}=\frac{4}{5} \Rightarrow q & =\frac{4}{5} \\
\therefore p=1-q & =1-\frac{4}{5}=\frac{1}{5} \\
n p=5 \Rightarrow n \times \frac{1}{5} & =5 \Rightarrow n=25 \\
\therefore(n, p) & \text { is }\left(25, \frac{1}{5}\right)
\end{aligned}
$

Question 10.
If the mean and standard deviation of a binomial distribution are 12 and 2 respectively. Then the value of its parameter $\mathrm{p}$ is ......
(a) $\frac{1}{2}$
(b) $\frac{1}{3}$
(c) $\frac{2}{3}$
(d) $\frac{1}{4}$
Solution:
(c) $\frac{2}{3}$
Hint:
$
\begin{aligned}
n p & =12 \\
\sqrt{n p q} & =2 \quad \Rightarrow \quad n p q=4 \\
\frac{n p q}{n p} & =\frac{4}{12}=\frac{1}{3} \therefore q=\frac{1}{3} \\
p & =1-q=1-\frac{1}{3}=\frac{2}{3}
\end{aligned}
$
Question 11.
A box contains 6 red and 4 white balls. If 3 balls are drawn at random, the probability of getting 2 white balls without replacement is
(a) $\frac{1}{20}$
(b) $\frac{18}{25}$
(c) $\frac{4}{25}$
(d) $\frac{3}{10}$
Solution:
(d) $\frac{3}{10}$
Hint:
Total number of balls $=10$
Number of ways of choosing 3 balls from 10 balls $=10 \mathrm{C}_3$
Number of ways of choosing 2 white balls from 4 white balls $=4 \mathrm{C}_3$
Number of ways of choosing 1 red ball from 6 red balls $=6 \mathrm{C}_1$
Probability of getting 2 white balls and 1 red ball $=\frac{6 C_1 \times 4 C_2}{10 C_3}$

$
=\frac{6 \times \frac{4 \times 3}{1 \times 2}}{\frac{10 \times 9 \times 8}{1 \times 2 \times 3}}=\frac{6 \times 6}{120}=\frac{3}{10}
$
Question 12.
If 2 cards are drawn from a well shuffled pack of 52 cards, the probability that they are of the same colours without replacement is ........
(a) $\frac{1}{2}$
(b) $\frac{26}{51}$
(c) $\frac{25}{51}$
(d) $\frac{25}{102}$
Solution:
(c) $\frac{25}{51}$
Hint:

Total number of cards $=52$
Number of ways of choosing
2 cards from 52 cards $]=52 \mathrm{C}_2$
Number of ways of choosing 2
Black cards from 26 Black cards $\}=26 \mathrm{C}_2$
Number of ways of choosing 2 ?
red cards from 26 red cards $\}=26 \mathrm{C}_2$
Probability of getting 2 cards $]=$ Probability of getting 2 black cards from
of same colour 26 black cards (or) Probability of getting 2 red cards from 26 red cards.
$
\begin{aligned}
& =\frac{26 \mathrm{C}_2+26 \mathrm{C}_2}{52 \mathrm{C}_2}=\frac{\frac{26 \times 25}{1 \times 2}+\frac{26 \times 25}{1 \times 2}}{\frac{52 \times 51}{1 \times 2}} \\
& =\frac{26 \times 25+26 \times 25}{52 \times 51}=\frac{2 \times 26 \times 25}{52 \times 51}=\frac{25}{51}
\end{aligned}
$
Question 13.
The distribution function $\mathrm{F}(\mathrm{X})$ of a random variable $\mathrm{X}$ is .......
(a) a decreasing function

(b) a non-decreasing function
(c) a constant function
(d) increasing first and then decreasing
Solution:
(b) a non-decreasing function