There are two bags. Bag-1 contains 4 white and 6 black balls and Bag-2 contains 5 white and 5 black balls. A die is rolled, if it shows a number divisible by 3, a ball is drawn from Bag-1, else a ball is drawn from Bag-2. If the ball drawn is not black in colour, the probability that it was not drawn from Bag-2 is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{2}{7}\)
Let \(E_1\): the ball is drawn from Bag-1, \(E_2\): the ball is drawn from Bag-2, and W: the ball is white. A ball that is “not black” is white, and “not from Bag-2” means from Bag-1, so the question asks for \(P(E_1 \mid W)\).
The numbers divisible by 3 on a die are 3 and 6:
\( \displaystyle P(E_1) = \frac{2}{6} = \frac{1}{3},\quad P(E_2) = \frac{2}{3} \)
\( \displaystyle P(W \mid E_1) = \frac{4}{10} = \frac{2}{5},\quad P(W \mid E_2) = \frac{5}{10} = \frac{1}{2} \)
By Bayes’ theorem:
\( \displaystyle P(E_1 \mid W) = \frac{P(E_1)P(W \mid E_1)}{P(E_1)P(W \mid E_1) + P(E_2)P(W \mid E_2)} \)
\( \displaystyle = \frac{\frac{1}{3}\times\frac{2}{5}}{\frac{1}{3}\times\frac{2}{5} + \frac{2}{3}\times\frac{1}{2}} \)
\( \displaystyle = \frac{\frac{2}{15}}{\frac{2}{15} + \frac{5}{15}} \)
\( \displaystyle = \frac{2}{7} \)