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