A bag contains 10 white and 15 black balls. If two balls are drawn in…

Consider the following events:

A = Getting a white ball in first draw,

B = Getting a black ball in second draw.

Required probability = Probability of getting a white ball in first draw and a black ball in second draw

⇒ Required probability = $P(A ∩ B)$

⇒ Required probability = P(A) P(B/A) .....(i)

Now,

$P(A) =\frac{^{10}C_1}{^{25}C_1}=\frac{10}{25}=\frac{2}{5}$

and, 

P(B/A) = Probability of getting a black ball in second draw when a white ball has already been in first draw

$⇒ P(B/A) =\frac{^{15}C_1}{^{24}C_1}=\frac{15}{24}=\frac{5}{8}$

∴ Required probability =$\frac{2}{5}× \frac{5}{8}=\frac{1}{4}$