Question

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 while the other ball turns out to be red, then the probability that both balls are drawn from box 'B' is :

• A. \(\frac{9}{16}\)
• B. \(\frac{7}{16}\)
• C. \(\frac{9}{32}\)
• D. \(\frac{7}{8}\)

Hint:

Use Bayes' theorem with equal prior probability 1/2 for each box.

Answer 1

The Correct Option is B

Let E = (one white and one red ball is drawn).
P(A) = P(B) = \(\frac{1}{2}\).
Box A has 7 balls: \(P(E|A) = \frac{2\cdot 3}{\binom{7}{2}} = \frac{6}{21} = \frac{2}{7}\).
Box B has 9 balls: \(P(E|B) = \frac{4\cdot 2}{\binom{9}{2}} = \frac{8}{36} = \frac{2}{9}\).
By Bayes' theorem: \(P(B|E) = \frac{\frac12\cdot\frac29}{\frac12\cdot\frac27+\frac12\cdot\frac29} = \frac{2/9}{2/7+2/9} = \frac{2/9}{32/63} = \frac{14}{32} = \frac{7}{16}\).
Answer: (2) \(\frac{7}{16}\)