Exercise 11.4-Additional Problems - Chapter 11 Probability Distributions 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
The probability of success of an event is $\mathrm{p}$ and that of failure is $q$. Find the expected number of trials to get a first success.
Solution:
Let $\mathrm{X}$ be the random variable denoting 'Number of trials to get a first success'. The success can occur in the $1^{\text {st }}$ trial. $\therefore$ The probability of success in the 1 st trial is $\mathrm{p}$. The success in the 2 nd trial means failure in the 1 st trial. $\therefore$ Probability is qp.
Success in the $3^{\text {rd }}$ trial means failure in the first two trials. Probability of success in the $3^{\text {rd }}$ trial is $\mathrm{q}^2 \mathrm{p}$. As it goes on, the success may occur in the $\mathrm{n}^{\text {th }}$ trial which mean the first $(\mathrm{n}-1)$ trials are failures. probability $=q^{n-1} p$.
$\therefore$ The probability distribution is as follows

$\begin{aligned}
\therefore \mathrm{E}(\mathrm{X}) & =\Sigma p_i x_i \\
& =1 . p+2 q p+3 q^2 p+\ldots+n q^{n-1} p . . \\
& =p\left[1+2 q+3 q^2+\ldots+n q^{n-1}+\ldots\right] \\
& =p[1-q]^{-2}=p(p)^{-2}=\frac{p}{p^2}=\frac{1}{p}
\end{aligned}$

Question 2.
An urn contains 4 white and 3 Red balls. Find the probability distribution of the number of red balls in three draws when a ball is drawn at random with replacement. Also find its mean and variance.
Solution:
The required probability distribution is

$
\text { Mean } \begin{aligned}
\mathrm{E}(\mathrm{X}) & =\Sigma p_i x_i \\
& =0\left(\frac{64}{343}\right)+1\left(\frac{144}{343}\right)+2\left(\frac{108}{343}\right)+3\left(\frac{27}{343}\right)=\frac{9}{7}
\end{aligned}
$
Variance $=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2$
$
\mathrm{E}\left(\mathrm{X}^2\right)=\Sigma p_i x_i^2
$
$
=0\left(\frac{64}{343}\right)+1^2\left(\frac{144}{343}\right)+2^2\left(\frac{108}{343}\right)+3^2\left(\frac{27}{343}\right)=\frac{117}{49}
$
$
\text { Variance }=\frac{117}{49}-\left(\frac{9}{7}\right)^2=\frac{36}{49}
$

Question 3.
$
f(x)=\left\{\begin{array}{cc}
3 e^{-3 x}, & 0 0, & \text { elsewhere }
\end{array}\right.
$
Find the mean and variance of the distribution

Solution:

$
\begin{aligned}
\mathrm{E}(\mathrm{X}) & =\int_{-\infty}^{\infty} x f(x) d x \\
& =\int_0^{\infty} x\left(3 e^{-3 x}\right) d x=3 \int_0^{\infty} x e^{-3 x} d x=3 \cdot \frac{\lfloor}{3^2}=\frac{1}{3} \\
\mathrm{E}\left(\mathrm{X}^2\right) & =\int_0^{\infty} x^2\left(3 e^{-3 x}\right) d x=3 \int_0^{\infty} x^2 e^{-3 x} d x=3 \cdot \frac{2}{3^3}=\frac{2}{9} \\
\operatorname{Var}(\mathrm{X}) & =\mathrm{E}\left[\mathrm{X}^2\right]-\mathrm{E}[\mathrm{X}]^2=\frac{2}{9}-\left(\frac{1}{3}\right)^2=\frac{1}{9} \\
\therefore \text { Mean } & =\frac{1}{3} ; \text { Variance }=\frac{1}{9}
\end{aligned}
$

Question 4.
Two cards are drawn with replacement from a well shuffled deck of 52 cards. Find the mean and variance for the number of aces.
Solution:
$\mathrm{n}(\mathrm{S})=52$
Number of aces $=n(A)=4$
$
\therefore P(A)=\frac{4}{52}=\frac{1}{13} . \text { So, } P\left(A^{\prime}\right)=1-\frac{1}{13}=\frac{12}{13}
$
Let $\mathrm{X}$ denotes number of aces when two cards are drawn.
$
\begin{aligned}
& P(X=0)=P\left(A^{\prime} A^{\prime}\right)=\frac{12}{13} \times \frac{12}{13}=\frac{144}{169} \\
& P(X=1)=P\left(A^{\prime}\right) \times 2 !=\frac{1}{13} \times \frac{12}{13} \times 2=\frac{24}{169} \\
& P(X=2)=P(A A)=\frac{1}{13} \times \frac{1}{13}=\frac{1}{169}
\end{aligned}
$
So, the probability distribution is

$\text { Mean } \begin{aligned}
\mathrm{E}(\mathrm{X}) & =\Sigma x_i p_i=0\left(\frac{144}{169}\right)+1\left(\frac{24}{169}\right)+2\left(\frac{1}{169}\right) \\
& =0+\frac{24}{169}+\frac{2}{169}=\frac{26}{169}=\frac{2}{13} \\
\mathrm{E}\left(\mathrm{X}^2\right) & =\Sigma x_i^2 p_i=0\left(\frac{144}{169}\right)+1^2\left(\frac{24}{169}\right)+2^2\left(\frac{1}{169}\right) \\
& =0+\frac{24}{169}+\frac{4}{169}=\frac{28}{169} \\
\operatorname{Var}(\mathrm{X}) & =\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2=\frac{28}{169}-\left(\frac{2}{13}\right)^2=\frac{28}{169}-\frac{4}{169}=\frac{24}{169} \\
\text { Mean } & =\frac{2}{13} ; \text { Variance }=\frac{24}{169}
\end{aligned}$