An urn contains five balls. Two balls are drawn and are found to be white. If P is the probability that all the balls are white and P =a/b in simplest form then find a+ b

Let $A_i(i=1,2,3,4)$ be the event that the um contains 2, 3, 4 or 5 white balls and B the event that two white balls are drawn

We have to find $\mathrm{P}\left({\mathrm{A}}_4/\mathrm{B}\right)$

Since the four events ${\mathrm{A}}_1,{\mathrm{A}}_2,{\mathrm{A}}_3,{\mathrm{A}}_4$ re equally likely, we have 

$\mathrm{P}\left({\mathrm{A}}_1\right)=\mathrm{P}\left({\mathrm{A}}_2\right)=\mathrm{P}\left({\mathrm{A}}_3\right)=\mathrm{P}\left({\mathrm{A}}_4\right)=\frac{1}{4}$

$\mathrm{P}\left(\mathrm{B}/{\mathrm{A}}_1\right)$ is the probability of event that the um contains 2 white balls and both have been drawn.

$$\begin{gathered} \text{ Hence }\mathrm{P}\left(\mathrm{B}/{\mathrm{A}}_1\right)=\frac{{ }^2{C}_2}{{ }^5{C}_2}=\frac{1}{10} \\ \text{ Similarly }\mathrm{P}\left(\mathrm{B}/{\mathrm{A}}_2\right)=\frac{{ }^3{C}_2}{{ }^5{C}_2}=\frac{3}{10} \\ \mathrm{P}\left(\mathrm{B}/{\mathrm{A}}_3\right)=\frac{{ }^4{C}_2}{{ }^5{C}_2}=\frac{6}{10}=\frac{3}{5}\text{ and }\mathrm{P}\left(\mathrm{B}/{\mathrm{A}}_4\right)=\frac{{ }^5{C}_2}{{ }^5{C}_2}=1 \end{gathered}$$

by Baye's theorem

$\mathrm{P}\left({\mathrm{A}}_4/\mathrm{B}\right)=\frac{P\left(A_4\right)P\left(B/A_4\right)}{\sum _{i=1}^4 P\left(A_1\right)P\left(B/A_i\right)}$

$=\frac{\frac{1}{4}\cdot 1}{\frac{1}{4}\left(\frac{1}{10}+\frac{3}{10}+\frac{3}{5}+1\right)}=\frac{1}{2}$

a=1 b=2

a+b=1+2=3