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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 drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

This question was previously asked in
HTET 2025 Level 1 PRT Question Paper (5-Jul-2026)
The correct answer is

$\frac{16}{31}$

The transferred ball is red with probability \(3/7\) (Bag I has 3 red out of 7) and black with probability \(4/7\).

If the transferred ball is red, Bag II has 5 red and 5 black balls (10 total), so \(P(\text{draw red}\mid\text{transferred red}) = 5/10=1/2\).

If the transferred ball is black, Bag II has 4 red and 6 black balls (10 total), so \(P(\text{draw red}\mid\text{transferred black}) = 4/10=2/5\).

By the law of total probability, \(P(\text{draw red}) = \left(\frac{3}{7}\right)\left(\frac{1}{2}\right)+\left(\frac{4}{7}\right)\left(\frac{2}{5}\right) = \frac{3}{14}+\frac{8}{35} = \frac{15}{70}+\frac{16}{70}=\frac{31}{70}\).

By Bayes' theorem, \(P(\text{transferred black}\mid\text{draw red}) = \dfrac{(4/7)(2/5)}{31/70} = \dfrac{8/35}{31/70} = \dfrac{16}{31}\).

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