A box A contains 2 white,3 red and 2 black balls. Another box B contains 4 white, 2 red and 3 black balls. If two balls are drawn at random, without replacement. from a randomly selected box and one ball turns out to be white the other balls turns out to be red, then the probability that both balls are drawn from box B is

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9/32

7/8

7/16

9/16

answer is A.

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Detailed Solution

Let A₁, A₂ be the events of selecting boxes A, B respectively. Then P(A₁) = 1/2, P(A₂) = 1/2.

Let E be the event of selecting one white and one red ball from the selected box.

Now $P (E ∣ A_{1}) = \frac{2 \times 3}{{}^{7}C_{2}} = \frac{2}{7}, P (E ∣ A_{2}) = \frac{4 \times 2}{{}^{9}C_{2}} = \frac{2}{9}$

Required probability $= P (A_{2} ∣ E) = \frac{P (A_{2}) P (E ∣ A_{2})}{P (A_{1}) P (E ∣ A_{1}) + P (A_{2}) P (E ∣ A_{2})} = \frac{\frac{1}{2} \times \frac{2}{9}}{\frac{1}{2} \times \frac{2}{7} + \frac{1}{2} \times \frac{2}{9}} = \frac{14}{18 + 14} = \frac{7}{16}$