Question

Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then then probability that it is drawn from urn A is :

• A. $\frac{4}{17}$
• B. $\frac{5}{18}$
• C. $\frac{7}{18}$
• D. $\frac{5}{16}$

Answer 1

Answer: (B)

$\frac{5}{18}$

Answer 2

Let $P(A) = \frac{1}{3}, \, P(B) = \frac{1}{3},$ and $P(C) = \frac{1}{3},$ since each urn is equally likely to be chosen.

Conditional Probabilities of Drawing a Black Ball:

$P(\text{Black}|A) = \frac{5}{12}, \quad P(\text{Black}|B) = \frac{7}{12}, \quad P(\text{Black}|C) = \frac{6}{12}$

Total Probability of Drawing a Black Ball:

$P(\text{Black}) = P(A) \times P(\text{Black}|A) + P(B) \times P(\text{Black}|B) + P(C) \times P(\text{Black}|C)$

$= \frac{1}{3} \times \frac{5}{12} + \frac{1}{3} \times \frac{7}{12} + \frac{1}{3} \times \frac{6}{12}$

$= \frac{18}{36} = \frac{1}{2}$

Using Bayes' Theorem:

$P(A|\text{Black}) = \frac{P(A) \times P(\text{Black}|A)}{P(\text{Black})} = \frac{\frac{1}{3} \times \frac{5}{12}}{\frac{1}{2}} = \frac{5}{18}$