JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Probability

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Bag A contains 2 white, 1 black and 3 red balls and bag B contains 3 black, 2 red and n white balls. One bag is chosen at random and 2 balls drawn from it at random, are found to be 1 red and 1 black. If the probability that both balls come from Bag A is

\frac{6}{11}

, then n is equal to

(A)13
(B)6
(C)4✓ Correct
(D)3

Answer: (C)

Solution

Let

E_{1}

= bag A chosen,

E_{2}

= bag B chosen,

A

= event (1 red, 1 black).

P (E_{1}) = P (E_{2}) = \frac{1}{2}

P (A ∣ E_{1}) = \frac{^{3} C _{1} \cdot ^{1} C _{1}}{^{6} C _{2}} = \frac{3}{15} = \frac{1}{5}

P (A ∣ E_{2}) = \frac{^{2} C _{1} \cdot ^{3} C _{1}}{^{n + 5} C _{2}} = \frac{6}{( n + 5 ) ( n + 4 ) /2} = \frac{12}{( n + 5 ) ( n + 4 )}

. By Bayes' theorem,

P (E_{1} ∣ A) = \frac{P ( E _{1} ) P ( A ∣ E _{1} )}{P ( E _{1} ) P ( A ∣ E _{1} ) + P ( E _{2} ) P ( A ∣ E _{2} )} = \frac{\frac{1}{10}}{\frac{1}{10} + \frac{6}{( n + 5 ) ( n + 4 )}} = \frac{6}{11}

. Solving gives

n = 4

JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Single correct option

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JEE Main · 2022 · 24 Jun 2022 · Shift 1 · Probability

Solution

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