Question

Match the follwoing

Answer 1

(A) Total number of ways $=6^5$

To find the favourable number of ways, a total of $12$ in $5$ throws can be obtained in the following two ways only

(i) One blank and four $3^{\prime }$s. or, (ii) Three $2^{\prime }$s and two $3^{\prime }$s.

The number of ways in case (i) ${=}^5{C}_1=5$

and the number of ways in case (ii) ${=}^5{C}_2=10.$

$\therefore m=$ the favourable number of ways.

$=5+10=15.$

Hence, the required probability $=\frac{15}{6^5}=\frac{5}{2592}.$

(B) Required probability

$=\frac{{}^{39}{C}_1}{{}^{52}{C}_1}\times \frac{{}^{39}{C}_1}{{}^{52}{C}_1}\times \frac{{}^{13}{C}_1}{{}^{52}{C}_1}=\frac{3}{4}\times \frac{3}{4}\times \frac{1}{4}=\frac{9}{64}.$

(C) Let $W$ denote the event that $A$ draws a white ball and $T$ the event that $A$ speaks truth.

In the usual notations, we are given that $P\left(W\right)=\frac{1}{9},P\left(\frac{T}{W}\right)=\frac{5}{6}$ so that $P\left(\overline{W}\right)=1-\frac{1}{9}=\frac{8}{9}$ and $P\left(\frac{T}{\overline{W}}\right)=1-\frac{5}{6}=\frac{1}{6}.$

Using Baye's theorem required probability is given by

$P\left(\frac{W}{T}\right)=\frac{P(W\cap T)}{P\left(T\right)}$ $=\frac{P\left(W\right)\times P\left(\frac{T}{W}\right)}{P\left(W\right)\times P\left(\frac{T}{W}\right)+P\left(\overline{W}\right)\times P\left(\frac{T}{\overline{W}}\right)}$ $=\frac{\left(\frac{1}{9}\right)\times \left(\frac{5}{6}\right)}{\left(\frac{1}{9}\right)\times \left(\frac{5}{6}\right)+\left(\frac{8}{9}\right)\times \left(\frac{1}{6}\right)}=\frac{5}{13}.$