A bag contains 5 white and 3 black balls; another bag contains 4 white and 5 black balls. From any one of these bags a single draw of two balls is made. Find the probability that one of them would be white and another black ball.

 Let P(B₁) and P(B₂) denote the probabilities of the events of drawing balls from the bag I or bag II.

Since both the bags are equally likely to be selected, P(B₁) = P(B₂) = $$\ \frac{\ 1}{2}$$

Let E denote the event of drawing two balls of different colours from a bag.

The required probability

= P(B₁)P($$\ \frac{\ E}{B_1}$$)+ P(B₂)P($$\ \frac{\ E}{B_2}$$)

P($$\ \frac{\ E}{B_1}$$) denote the probability of drawing the two balls from bag I and as that from bag II.

= $$\ \frac{\ 1}{2}\left[\ \frac{\ ^5C_1\ \times\ \ ^3C_1}{^8C_2}\ +\ \ \frac{\ ^5C_1\ \times\ \ ^4C_1}{^9C_2}\right]$$

= $$\ \frac{\ 1}{2}\left[\ \frac{\ 15}{28}+\ \frac{\ 5}{9}\right]$$

= $$\ \frac{\ 275}{504}$$

A is the correct answer.