Refer to question 41 above. If a white ball is selected, what is the probability that it came from

(i) Bag II?

(ii) Bag III?

Referring to the previous Solution, using Bay's theorem, we have

(i) $P\left( {{E_2}/F} \right) = \frac{{P\left( {{E_2}} \right) \cdot P\left( {F/{E_2}} \right)}}{{P\left( {{E_1}} \right) \cdot P\left( {F/{E_1}} \right) + P\left( {{E_2}} \right) \cdot P\left( {F/{E_2}} \right) + P\left( {{E_3}} \right) \cdot P\left( {F/{E_3}} \right)}}$

$= \frac{{\frac{2}{6} \cdot \frac{1}{3}}}{{\frac{1}{6} \cdot 0 + \frac{2}{6} \cdot \frac{1}{3} + \frac{3}{6} \cdot 1}} = \frac{{\frac{2}{{18}}}}{{\frac{2}{{18}} + \frac{3}{6}}}$

$= \frac{{2/18}}{{\frac{{2 + 9}}{{18}}}} = \frac{2}{{11}}$

(ii) $P\left( {{E_3}/F} \right) = \frac{{P\left( {{E_3}} \right) \cdot P\left( {F/{E_3}} \right)}}{{P\left( {{E_1}} \right) \cdot P\left( {F/{E_1}} \right) + P\left( {{E_2}} \right) \cdot P\left( {F/{E_2}} \right) + P\left( {{E_3}} \right) \cdot P\left( {F/{E_3}} \right)}}$

$= \frac{{\frac{3}{6} \cdot 1}}{{\frac{1}{6} \cdot 0 + \frac{2}{6} \cdot \frac{1}{3} + \frac{3}{6} \cdot 1}}$

$= \frac{{\frac{3}{6}}}{{\frac{2}{{18}} + \frac{3}{6}}} = \frac{{3/6}}{{\frac{2}{{18}} + \frac{9}{{18}}}} = \frac{9}{{11}}$