Question

A bag contains 7 red and 2 white balls and another bag contains 5 red and 4 white balls. Two balls are drawn, one from each bag. The probability that both the balls are white, is

• A. $\frac{2}{9}$
• B. $\frac{2}{3}$
• C. $\frac{8}{81}$
• D. $\frac{35}{81}$

Answer 1

The correct option is C $\frac{8}{81}$

Both the bags contain 9 balls each. We have to select one ball each from both of these boxes out of the 9 balls available. That can be done in ${}^9{C}_1$ ways each.
For the favorable outcomes, we have to select 1 white ball from the first box out of the 2 white balls and 1 from second balls which has 4 white balls.
Therefore required probability =$\frac{{}^2{C}_1}{{}^9{C}_1}\times \frac{{}^4{C}_1}{{}^9{C}_1}$
$=\frac{8}{81}$