$\text{A bag contains 5 balls of unknown colors such that a ball is drawn at random from it and is found to be white.}$

$\mathrm{The} \mathrm{probability} \mathrm{that} \mathrm{bag} \mathrm{contains} \mathrm{only} \mathrm{white} \mathrm{balls} \mathrm{is}$

$\text{ Let }{\mathrm{W}}_{\mathrm{i}}(\mathrm{i}=1,2,3,4,5)\text{ denotes }\mathrm{i}\text{ white balls and }(5-i)\text{ denotes }$

 other coloured balls.

 W denotes event of selecting white ball.

$\begin{array}{r}\mathrm{P}\left(\frac{\mathrm{W}}{{\mathrm{W}}_{\mathrm{i}}}\right)=\frac{\mathrm{i}}{5},(\mathrm{i}=1,2,3,4,5)\\ \mathrm{P}\left({\mathrm{W}}_{\mathrm{i}}\right)=\frac{1}{5},(\mathrm{i}=1,2,3,4,5)\end{array}$

Probability of occurrence of W given $W_5$  has occurred

$\begin{array}{l}\mathrm{P}\left(\frac{\mathrm{W}}{{\mathrm{W}}_5}\right)=\frac{\mathrm{P}\left({\mathrm{W}}_5\right)\cdot \mathrm{P}\left(\frac{\mathrm{W}}{{\mathrm{W}}_5}\right)}{\sum _{\mathrm{i}=1}^5 \mathrm{P}\left({\mathrm{W}}_{\mathrm{i}}\right)\mathrm{P}\left(\frac{\mathrm{W}}{{\mathrm{W}}_{\mathrm{i}}}\right)}\\ =\frac{\frac{1}{5}\times 1}{\frac{1}{5}\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}+\frac{5}{5}\right)}=\frac{1}{3}\\ \text{ Therefore, the correct answer is }0.33\end{array}$        $\left(Baye\text{'}s theorem\right)$