Urn A contains 1 white, 2 black and 3 red balls, Urn B contains 2 white, 1 black and 1 red ball and Urn C contains 4 white, 5 black and 3 red balls. One Urn is chosen at random and two balls are drawn. If it is found that there is one white and one red, then the probability that they come from A is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{33}{118} \)
This is Bayes’ theorem. Let E be the event “one white and one red ball”. Each urn is chosen with probability \( \frac{1}{3} \).
Probability of E from each urn (one white chosen from the whites and one red from the reds):
\( \displaystyle P(E|A)=\frac{1\times 3}{^{6}C_{2}}=\frac{3}{15}=\frac{1}{5} \)
\( \displaystyle P(E|B)=\frac{2\times 1}{^{4}C_{2}}=\frac{2}{6}=\frac{1}{3} \)
\( \displaystyle P(E|C)=\frac{4\times 3}{^{12}C_{2}}=\frac{12}{66}=\frac{2}{11} \)
Since the urns are equally likely, the factor \( \frac{1}{3} \) cancels:
\( \displaystyle P(A|E)=\frac{\frac{1}{5}}{\frac{1}{5}+\frac{1}{3}+\frac{2}{11}} \)
Over the common denominator 165, the three probabilities are \( \frac{33}{165} \), \( \frac{55}{165} \), \( \frac{30}{165} \):
\( \displaystyle P(A|E)=\frac{33}{33+55+30} \)
\( \displaystyle =\frac{33}{118} \)