Question

Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Answer 1

Bag I contains 3 red and 4 black balls.

Bag II contains 4 red and 5 black balls.

Let E₁ and E₂ be the events of drawing a red ball and a black ball from bag I, then

∴ P(E₁) = `3/7`, P(E₂) = `4/7`

Event A: Drawing a red ball

A red ball is taken out from bag I and placed in bag II. Thus bag II has 5 red and 5 black balls.

∴ `P(A/E_1) = 5/10`

A black ball is taken out from bag I and placed in bag II. Thus bag II contains 4 red and 6 black balls.

∴ `P(A/E_2) = 4/10`

By Bayes' theorem,

`P(E_2/E) = (P(E_2) xx P(E/E_2))/(P(E_1) xx P(E/E_1) + P(E_2) xx P(E/E_2))`

= `(4/7 xx 4/10)/(3/7 xx 5/10 + 4/7 xx 4/10)`

= `16/(15 + 16)`

= `16/31`