Question

A box B1 contains 1 white ball, 3 red balls, and 2 black balls. Another box B2 contains 2 white balls, 3 red balls, and 4 black balls. A third box B3 contains 3 white balls, 4 red balls, and 5 black balls. If 2 balls are drawn (without replacement) from a randomly selected box and one of the balls is white and the other ball is red, the probability that these 2 balls are drawn from box B2 is (a) \(\frac{116}{181}\) (b) \(\frac{126}{181}\) (c) \(\frac{65}{181}\) d) \(\frac{55}{181}\) Select the correct answer from above options

Answer 1

Answer: (D) = \(\frac{55}{181}\) Let E : Event of selecting red and one white ball Probability of selecting a box = P(B1) = P(B2) = P(B2) = \(\frac{1}{3}\) Probability of selecting 1 Red and 1 White ball from box B1 = \(P\big(\frac{E}{B_1}\big) = \frac{^1C_1 \times ^3C_1}{^6C_2} = \frac{1 \times 3 \times 2}{6 \times 5} = \frac{1}{5}\) = \(P\big(\frac{E}{B_2}\big) = \frac{^2C_1 \times ^3C_1}{^9C_2} = \frac{2 \times 3 \times 3}{9 \times 8} = \frac{1}{6}\) = \(P\big(\frac{E}{B_3}\big) = \frac{^3C_1 \times ^4C_1}{^{12}C_2} = \frac{3 \times 4 \times 2}{12 \times 11} = \frac{2}{11}\) \(\therefore\) \(P\big(\frac{B_2}{E}\big) = \frac {P(B_2) \times P\big(\frac{E}{B_2}\big)}{P(B_1) \times P\big(\frac{E}{B_1}\big) +P(B_2) \times P\big(\frac{E}{B_2}\big) +P(B_3) \times P\big(\frac{E}{B_3}\big)}\) (Using Bayes Theorem) = \(\frac{\frac{1}{3} \times \frac{1}{6}}{\frac{1}{3} \times \frac{1}{5} +\frac{1}{3} \times \frac{1}{6} +\frac{1}{3} \times \frac{2}{11}}\) = \(\frac {\frac{1}{6}}{\frac{66+55+30}{330}}\) = \(\frac{\frac{1}{6}}{\frac{181}{330}}\) = \(\frac{55}{181}\)