JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Probability

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Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white 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 black in colour. Then the probability, that the transferred ball is red, is:

(A) $\frac{4}{9}$
(B) $\frac{5}{18}$ ✓ Correct
(C) $\frac{1}{6}$
(D) $\frac{3}{10}$

Answer: (B)

Solution

Let

A

be the event that the drawn ball from Bag II is black, and

B

be the event that the transferred ball is red. By Bayes' theorem,

P (\frac{B}{A}) = \frac{P ( A \cap B )}{P ( A )}

. We have

P (A \cap B) = \frac{3}{10} \times \frac{5}{10} = \frac{15}{100}

(transfer red, then draw black).

P (A) = \frac{3}{10} \times \frac{5}{10} + \frac{4}{10} \times \frac{6}{10} + \frac{3}{10} \times \frac{5}{10} = \frac{15 + 24 + 15}{100} = \frac{54}{100}

. Thus

P (\frac{B}{A}) = \frac{15}{54} = \frac{5}{18}

JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 29 Jul 2022 · Shift 2 · Probability

Solution

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