Question

Three urn A, B, C, A has 7 red and 5 black balls, B has 5 red and 7 black balls, C has 6 red and 6 black balls. One urn is selected and black ball is taken out. Find probability that the selected urn is A.

• A. 7/18
• B. 5/17
• C. 7/19
• D. 5/18

Answer 1

The Correct Option is D

To solve this problem, we need to apply Bayes' theorem. The theorem helps in determining the probability of an event based on prior knowledge of conditions related to the event.

Given:

• Urn A: 7 red and 5 black balls
• Urn B: 5 red and 7 black balls
• Urn C: 6 red and 6 black balls

The problem asks for the probability that urn A was selected given that a black ball was drawn.

Let's define the events:

• U_A: Urn A is selected
• U_B: Urn B is selected
• U_C: Urn C is selected
• B: A black ball is drawn

We want to find P(U_A \mid B).

Using Bayes’ theorem:

P(U_A \mid B) = \frac{P(B \mid U_A) \cdot P(U_A)}{P(B)}

Since each urn is equally likely to be selected:

P(U_A) = P(U_B) = P(U_C) = \frac{1}{3}

Now, let's find P(B \mid U_A), P(B \mid U_B), and P(B \mid U_C):

• P(B \mid U_A) = \frac{5}{12} since there are 5 black balls out of 12 in urn A.
• P(B \mid U_B) = \frac{7}{12} since there are 7 black balls out of 12 in urn B.
• P(B \mid U_C) = \frac{6}{12} = \frac{1}{2} since there are 6 black balls out of 12 in urn C.

The total probability of drawing a black ball, P(B), is given by:

P(B) = P(B \mid U_A) \cdot P(U_A) + P(B \mid U_B) \cdot P(U_B) + P(B \mid U_C) \cdot P(U_C)

P(B) = \frac{5}{12} \cdot \frac{1}{3} + \frac{7}{12} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3}

P(B) = \frac{5}{36} + \frac{7}{36} + \frac{6}{36} = \frac{18}{36} = \frac{1}{2}

Finally, using Bayes’ theorem, find P(U_A \mid B):

P(U_A \mid B) = \frac{\frac{5}{12} \cdot \frac{1}{3}}{\frac{1}{2}}

P(U_A \mid B) = \frac{5}{12} \cdot \frac{1}{3} \cdot \frac{2}{1} = \frac{5}{18}

Thus, the probability that the selected urn is A, given that a black ball was drawn, is \frac{5}{18}. Therefore, the correct answer is 5/18.