There are two bags, I and II. Bag I contain 3 and 5 black balls and bag II contains 4 red and 3 black balls. One ball is transferred randomly from Bag I to Bag II and then a ball is drawn randomly from Bag II. If the ball so drawn is found to be black in colour, then find the probability that the transferred ball is also black.

OR

An urn contain 5 red, 2 white and 3 black balls. Three balls are drawn, one-by-one, at random without replacement. Find the probability distribution of the number of white balls. Also, find the mean and the variance of the number of white balls drawn.

Let,
A= Ball transferred from Bag I to Bag II is black
B= Ball transferred from Bag I to Bag II is red
C= Ball transferred from Bag II is black.
$\begin{array}{r}\therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{5}{8}\\ \therefore \mathrm{P}\left(\mathrm{B}\right)=\frac{3}{8}\end{array}$
Now,
The probability that black ball is drawn from bag II, if black ball is transferred from Bag I.
$\mathrm{P}(\mathrm{C}\mid \mathrm{A})=\frac{4}{8}=\frac{1}{2}$
The probability that black ball is drawn from bag II, if red ball is transferred from Bag I.
$\mathrm{P}(\mathrm{C}\mid \mathrm{B})=\frac{3}{8}$
Now, we need to find out the probability that the ball transferred is black, if ball drawn is black in colour,
$\mathrm{P}(\mathrm{A}\mid \mathrm{C})=\frac{\mathrm{P}\left(\mathrm{A}\right)\cdot \mathrm{P}(\mathrm{C}/\mathrm{A})}{\mathrm{P}\left(\mathrm{A}\right)\cdot \mathrm{P}(\mathrm{C}/\mathrm{A})+\mathrm{P}\left(\mathrm{B}\right)\cdot \mathrm{P}(\mathrm{C}/\mathrm{B})}=\frac{\frac{51}{82}}{\frac{5}{8}\frac{1}{2}+\frac{3}{8}\cdot \frac{3}{8}}=\frac{20}{29}$
Which is the answer.

OR

Let assume that the number of white balls are X. Notice the below table,

Now, using the following formula to find the Mean,
Mean$=\mathrm{\Sigma X}\cdot \mathrm{P}\left(\mathrm{X}\right)=\frac{7}{15}+\frac{2}{15}=\frac{3}{5}$
Now, using the following formula to find the Variance,
Variance $={\mathrm{\Sigma X}}^2\cdot \mathrm{P}\left(\mathrm{X}\right)-\left[\mathrm{\Sigma XP}\right(\mathrm{X}){]}^2=\frac{7}{15}+\frac{4}{15}-{\left(\frac{3}{5}\right)}^2=\frac{11}{15}-\frac{9}{25}=\frac{28}{75}$
Which are the answers.