PiyushK wrote:

A bag contains 3 red and 2 black ball. Another bag contains 4 red and 5 black balls. A ball is drawn from the first bag and is placed in the second. A ball is then drawn from the second. What is the probability that this draw of red ball is due to the drawing of red ball from the first bag ?

A. 3/5

B. 2/5

C. 4/25

D. 3/10

E. 15/23

Given:

Bag 1: 3 Red, 2 Black

Bag 2: 4 Red, 5 Black

1st transaction:

A ball is moved from bag 1 and shifted to bag 2

P(R) = \(\frac{3}{5}\) and P(B) = \(\frac{2}{5}\)

2nd Transaction:

A ball is picked from the Bag 2.

Required: Probability that this draw of red ball is due to the drawing of red ball from the first bag

Case 1: Black ball was picked from Bag 1 and then a Red is picked from Bag 2

P(R) = \(\frac{2}{5}*\frac{4}{10} = \frac{8}{50}\)

Case 2: Red ball was picked from Bag 1 and then a Red is picked from Bag 2

P(R) =\(\frac{3}{5}*\frac{5}{10}= \frac{15}{50}\)

We need the probability of occurrence off Case 2:

Hence P(Case 2) =\(\frac{15}{50} / (\frac{8}{50} + \frac{15}{50}) = \frac{15}{23}\)

Option E