First bag contains 5 white and 4 black balls. Second bag contains 7 white and 9 black balls. A ball is transferred from the first bag to the second bag and then a ball is drawn from the second bag. Find the probability that the ball drawn is white. 

Given:

First bag contains 5 white and 4 black balls.

Second bag contains 7 white and 9 black balls.

A ball is transferred from the first bag to the second bag and then a ball is drawn from the second bag.

Formula used:

Probability of drawing a white ball from the second bag = P(transfer white) × P(draw white after transfer white) + P(transfer black) × P(draw white after transfer black)

Calculations:

Probability of transferring a white ball from the first bag: \(\dfrac{5}{9}\)

Probability of transferring a black ball from the first bag: \(\dfrac{4}{9}\)

After transferring a white ball:

Second bag contains 8 white and 9 black balls.

Probability of drawing a white ball from the second bag: \(\dfrac{8}{17}\)

After transferring a black ball:

Second bag contains 7 white and 10 black balls.

Probability of drawing a white ball from the second bag: \(\dfrac{7}{17}\)

⇒ Total probability of drawing a white ball from the second bag:

⇒ \(\dfrac{5}{9} × \dfrac{8}{17} + \dfrac{4}{9} × \dfrac{7}{17}\)

⇒ \(\dfrac{5 × 8}{9 × 17} + \dfrac{4 × 7}{9 × 17}\)

⇒ \(\dfrac{40}{153} + \dfrac{28}{153}\)

⇒ \(\dfrac{68}{153}\)

⇒ \(\dfrac{4}{9}\)

∴ The correct answer is option (2).