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JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Probability

If A and B are two events such that

P (A) = \frac{1}{3}, P (B) = \frac{1}{5}

and

P (A \cup B) = \frac{1}{2}

, then

P (A ∣ B^{'}) + P (B ∣ A^{'})

is equal to

(A) $\frac{3}{4}$
(B) $\frac{5}{8}$ ✓ Correct
(C) $\frac{5}{4}$
(D) $\frac{7}{8}$

Answer: (B)

Solution

First find

P (A \cap B) = P (A) + P (B) - P (A \cup B) = \frac{1}{3} + \frac{1}{5} - \frac{1}{2} = \frac{10 + 6 - 15}{30} = \frac{1}{30}

. Then

P (A ∣ B^{'}) = \frac{P ( A \cap B ^{'} )}{P ( B ^{'} )} = \frac{P ( A ) - P ( A \cap B )}{1 - P ( B )} = \frac{\frac{1}{3} - \frac{1}{30}}{\frac{4}{5}} = \frac{\frac{9}{30}}{\frac{4}{5}} = \frac{9}{30} \times \frac{5}{4} = \frac{3}{8}

. Similarly,

P (B ∣ A^{'}) = \frac{P ( B \cap A ^{'} )}{P ( A ^{'} )} = \frac{P ( B ) - P ( A \cap B )}{1 - P ( A )} = \frac{\frac{1}{5} - \frac{1}{30}}{\frac{2}{3}} = \frac{\frac{5}{30}}{\frac{2}{3}} = \frac{1}{6} \times \frac{3}{2} = \frac{1}{4}

. Therefore,

P (A ∣ B^{'}) + P (B ∣ A^{'}) = \frac{3}{8} + \frac{1}{4} = \frac{3}{8} + \frac{2}{8} = \frac{5}{8}

.

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Single correct option

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