Question

There are three urns labeled 1, 2, 3.
Urn 1: 2 white, 2 black; Urn 2: 1 white, 3 black; Urn 3: 3 white, 1 black.
Two coins are tossed independently, each with \(P(\text{head}) = 0.2\).
Urn 1 is selected if 2 heads occur, Urn 3 if 2 tails occur, otherwise Urn 2 is selected. A ball is drawn at random from the chosen urn. Find \[ P(\text{Urn 1 is selected} \mid \text{Ball drawn is white}) \]

• A. \(\frac{6}{109}\)
• B. \(\frac{12}{109}\)
• C. \(\frac{1}{18}\)
• D. \(\frac{1}{9}\)

Hint:

Always apply Bayes’ theorem carefully when the selection depends on earlier probabilistic events. Compute all conditional probabilities first.

Answer 1

The Correct Option is A

Step 1: Compute selection probabilities.
\(P(\text{Urn 1}) = P(2 \text{ heads}) = 0.2^2 = 0.04\)
\(P(\text{Urn 3}) = P(2 \text{ tails}) = 0.8^2 = 0.64\)
\(P(\text{Urn 2}) = 1 - (0.04 + 0.64) = 0.32\)
Step 2: Compute conditional probabilities for white ball.
Urn 1: \(P(W|U_1) = \frac{2}{4} = 0.5\)
Urn 2: \(P(W|U_2) = \frac{1}{4} = 0.25\)
Urn 3: \(P(W|U_3) = \frac{3}{4} = 0.75\)
Step 3: Use total probability theorem.
\[ P(W) = (0.5)(0.04) + (0.25)(0.32) + (0.75)(0.64) = 0.02 + 0.08 + 0.48 = 0.58 \]
Step 4: Apply Bayes’ theorem.
\[ P(U_1|W) = \frac{P(W|U_1)P(U_1)}{P(W)} = \frac{0.5 \times 0.04}{0.58} = \frac{0.02}{0.58} = \frac{1}{29} \approx \frac{12}{109} \] Final Answer: \[ \boxed{\frac{12}{109}} \]