P(\(E_1)=\large\frac{m}{m+n}\)
P(\(E_2)=\large\frac{n}{m+n}\)
Now if first ball is drawn is White'k' white ball are added Total balls are m+n+k
P(\(A/E-1\))=P(getting white in 2\(^{nd}\) / 1\(^{st}\) ball drawn is'w')
=\(\large\frac{m+k}{m+n+k}\)
P(\(A/E-2\))=P(getting white in 2\(^{nd}\) / 1\(^{st}\) ball drawn is'b')
=\(\large\frac{m}{m+n+k}\)
P(A)=\(\Large\frac{m}{m+n}\)\(\times\)\(\frac{m+k}{m+n+k}\)\(\times\)\(\frac{n}{m+n}\)\(\times\)\(\frac{m}{m+n+k}\)
=\(\Large\frac{m(m+n+k)}{(m+n)(m+n+k)}\)
=\(\large\frac{m}{m+n}\)
this probability is independent of 'k'