Question

Bag A contains 3 red and 2 white balls, and Bag B contains 2 red and 5 white balls. A bag selected at random, a ball is drawn and put into the other bag; then a ball is drawn from the second bag. Find the probability that both balls drawn are of the same colour.

Answer 1

Let $E_{11}$ be the event of transferring a red ball from Bag A to Bag B $=\frac{3}{5}$

Let $A$ be the event of drawing a red ball from Bag B $P\left(\frac{A}{E_{11}}\right)=\frac{3}{8}$

Let $E_{12}$ be the event of transferring a white ball from Bag A to Bag B $=\frac{2}{5}$

Let $B$ be the event of drawing a white ball from Bag B $P\left(\frac{B}{E_{12}}\right)=\frac{6}{8}$

Let $E_{21}$ be the event of transferring a red ball from Bag B to Bag A $=\frac{2}{7}$

Let $C$ be the event of drawing a red ball from Bag A $P\left(\frac{C}{E_{21}}\right)=\frac{4}{6}$

Let $E_{22}$ be the event of transferring a white ball from Bag B to Bag A $=\frac{5}{7}$

Let $D$ be the event of drawing a white ball from Bag A $P\left(\frac{D}{E_{22}}\right)=\frac{3}{6}$

Hence

Required $P\left(E_1\right)=P\left(E_{11}\right).P\left(\frac{A}{E_{11}}\right)+P\left(E_{12}\right).P\left(\frac{B}{E_{12}}\right)$

$=\left(\frac{3}{5}\right).\left(\frac{3}{8}\right)+\left(\frac{2}{5}\right)\left(\frac{6}{8}\right)$

$=\frac{9}{40}+\frac{12}{40}$

$=\frac{21}{40}$

Required $P\left(E_2\right)=P\left(E_{21}\right).P\left(\frac{C}{E_{21}}\right)+P\left(E_{22}\right).P\left(\frac{D}{E_{22}}\right)$

$=\left(\frac{2}{7}\right).\left(\frac{4}{6}\right)+\left(\frac{5}{7}\right)\left(\frac{3}{6}\right)$

$=\frac{8}{42}+\frac{15}{42}$

$=\frac{23}{42}$

$\therefore P\left(E\right)=0.5P\left(E_1\right)+0.5P\left(E_2\right)$ (since a bag was picked randomly with equal probability)

$P\left(E\right)=0.5(\frac{21}{40}+\frac{23}{42})=\frac{901}{1680}$