Exercise 11.2-Additional Problems - Chapter 11 - Probability Distributions - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Problems
Question 1.
Find the probability mass function, and the cumulative distribution function for getting ' 3 's when two dice are thrown.
Solution:
Two dice are thrown. Let $\mathrm{X}$ be the random variable of getting number of ' 3 's.
Therefore $\mathrm{X}$ can take the values $0,1,2$.
Sample Space
$
\begin{aligned}
& P\left(\text { no } 3^{\prime}\right)=P(X=0)=\frac{25}{36} \\
& P\left(\text { one }^{\prime} 3^{\prime}\right)=P(X=1)=\frac{10}{36} \\
& P\left(\text { two } 3^{\prime}\right)=P(X=2)=\frac{1}{36}
\end{aligned}
$
Probability mass function is given by


Cumulative distribution function:
We have $\mathrm{F}(x)=\sum_{x_i=-\infty}^x \mathrm{P}\left(\mathrm{X}=x_i\right)$
$
\begin{aligned}
& \mathrm{F}(0)=\mathrm{P}(\mathrm{X}=0)=\frac{25}{36} \\
& \mathrm{~F}(\mathrm{I})=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)=\frac{25}{36}+\frac{10}{36}=\frac{35}{36} \\
& \mathrm{~F}(2)=\mathrm{P}(\mathrm{X}=0)+\mathrm{P}(\mathrm{X}=1)+\mathrm{P}(\mathrm{X}=2)=\frac{25}{36}+\frac{10}{36}+\frac{1}{36}=\frac{36}{36}=1
\end{aligned}
$

Question 2.
If $F(x)=\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} x\right)-\infty

Solution:
$
\begin{aligned}
\mathrm{F}(x) & =\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} x\right) \\
\mathrm{P}(0 \leq x \leq 1) & =\mathrm{F}(1)-\mathrm{F}(0) \\
& =\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} 1\right)-\frac{1}{\pi}\left(\frac{\pi}{2}+\tan ^{-1} 0\right) \\
& =\frac{1}{\pi}\left(\frac{\pi}{2}+\frac{\pi}{4}\right)-\frac{1}{\pi}\left(\frac{\pi}{2}+0\right)=\frac{1}{\pi}\left(\frac{\pi}{2}+\frac{\pi}{4}-\frac{\pi}{2}\right)=\frac{1}{4}
\end{aligned}
$
Question 3.
If $f(x)=\left\{\begin{array}{ll}\frac{A}{x}, & 1e)$
Solution:
Since $f(x)$ is a probability density function, $\int_{-\infty}^{\infty} f(x) d x=1$
$
\begin{aligned}
& \int_1^{e^3} \frac{\mathrm{A}}{x} d x=1 \Rightarrow \mathrm{A}[\log x]_1^{e^3}=1 \\
& \Rightarrow \mathrm{A}\left[\log e^3-\log 1\right]=1 \Rightarrow \mathrm{A}[3]=1 \Rightarrow \mathrm{A}=1 / 3 \\
& \text { Therefore } f(x)= \begin{cases}\frac{1}{3 x}, & 1 0 & \text { elsewhere }\end{cases} \\
& \mathrm{P}(x>e)=\frac{1}{3} \int_e^{e^3} \frac{1}{x} d x=\frac{1}{3}[\log x]_e^{e^3} \\
& =\frac{1}{3}\left[\log e^3-\log e\right]=\frac{1}{3}[3-1]=\frac{2}{3} \\
&
\end{aligned}
$

Question 4.

(i) $\mathrm{P}(0.5<\mathrm{X}<0.75)$
(ii) $\mathrm{P}(\mathrm{X} \leq 0.5)$
(iii) $\mathrm{P}(\mathrm{X}>0.75)$

Solution:

(i) $\mathrm{P}(0.5<\mathrm{X}<0.75)=\int_{0.5}^{0.75} f(x) d x=\int_{0.5}^{0.75} 2 x d x=\left[x^2\right]_{0.5}^{0.75}=0.5625-0.25=0.3125$
(ii) $\mathrm{P}(\mathrm{X} \leq 0.5)=\int_0^{0.5} f(x) d x=\int_0^{0.5} 2 x d x=\left[x^2\right]_0^{0.5}=0.25-0=0.25$
(iii) $P(X>0.75)=\int_{0.75}^1 f(x) d x=\int_{0.75}^1 2 x d x=\left[x^2\right]_{0.75}^1=1-0.5625=0.4375$

Question 5.

Solution:

(i) Since $f(x)$ is a p.d.f. $\int_{-\infty} f(x) d x=1$
i.e.,
$
\int_0^{2 \pi} k d x=1[k x]_0^{2 \pi}=1 \text { i.e., } k\left[[2 \pi]=1 \Rightarrow k=\frac{1}{2 \pi}\right.
$

(ii) $\mathrm{P}(0<\mathrm{X}<\pi / 2)=\int_0^{\pi / 2} k d x=\frac{1}{2 \pi}[x]_0^{\pi / 2}=\frac{1}{2 \pi}\left[\frac{\pi}{2}-0\right]=\frac{1}{2 \pi}\left[\frac{\pi}{2}\right]=\frac{1}{4}$
(iii) $\mathrm{P}\left(\frac{\pi}{2}