Problem 9 - A First Course in Probability

Question

Consider 3 urns. Urn \(A\) contains 2 white and 4 red balls, urn \(B\) contains 8 white and 4 red balls, and urn \(C\) contains 1 white and 3 red balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from urn \(A\) was white given that exactly 2 white balls were selected?

Step-by-Step Solution

Answer

The probability that the ball chosen from urn A was white given that exactly 2 white balls were selected is \(\frac{7}{13}\).

Step 1: Identify the relevant events and probabilities

Let us define the events of interest: - Event A: A white ball is selected from urn A. - Event B: A white ball is selected from urn B. - Event C: A white ball is selected from urn C. - Event W: Exactly 2 white balls are selected in total. We are asked to find the probability of event A given event W, or \(P(A|W)\). First, we find the probabilities of selecting a white or red ball from each urn: - Urn A: \(P(A) = \frac{2}{6} = \frac{1}{3}\) and \(P(\overline{A}) = \frac{4}{6} = \frac{2}{3}\) - Urn B: \(P(B) = \frac{8}{12} = \frac{2}{3}\) and \(P(\overline{B}) = \frac{4}{12} = \frac{1}{3}\) - Urn C: \(P(C) = \frac{1}{4}\) and \(P(\overline{C}) = \frac{3}{4}\)

Step 2: Identify all combinations that can result in exactly 2 white balls

To find event W (exactly 2 white balls are selected), we check all possible combinations that satisfy this condition: 1. White from urn A, white from urn B, and red from urn C: \(A \cap B \cap \overline{C}\) 2. White from urn A, red from urn B, and white from urn C: \(A \cap \overline{B} \cap C\) 3. Red from urn A, white from urn B, and white from urn C: \(\overline{A} \cap B \cap C\)

Step 3: Calculate the probabilities of each combination in Step 2

Using the probabilities found in Step 1, we calculate the probabilities of the combinations in Step 2: 1. \(P(A \cap B \cap \overline{C}) = P(A) \times P(B) \times P(\overline{C}) = \frac{1}{3} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{6}\) 2. \(P(A \cap \overline{B} \cap C) = P(A) \times P(\overline{B}) \times P(C) = \frac{1}{3} \times \frac{1}{3} \times \frac{1}{4} = \frac{1}{36}\) 3. \(P(\overline{A} \cap B \cap C) = P(\overline{A}) \times P(B) \times P(C) = \frac{2}{3} \times \frac{2}{3} \times \frac{1}{4} = \frac{1}{9}\) Now, we can find the probability of event W (exactly 2 white balls are selected): \(P(W) = P(A \cap B \cap \overline{C}) + P(A \cap \overline{B} \cap C) + P(\overline{A} \cap B \cap C) = \frac{1}{6} + \frac{1}{36} + \frac{1}{9} = \frac{13}{36}\)

Step 4: Apply conditional probability formula and find the required probability

We now have everything we need to find the probability of event A given event W (a white ball is chosen from urn A given exactly 2 white balls are selected). Using the conditional probability formula: \(P(A|W) = \frac{P(A \cap W)}{P(W)} = \frac{P(A \cap B \cap \overline{C}) + P(A \cap \overline{B} \cap C)}{P(W)}\) Plugging in the probabilities calculated in Steps 3 and 4: \(P(A|W) = \frac{\frac{1}{6} + \frac{1}{36}}{\frac{13}{36}} = \frac{7}{13}\) So the probability that the ball chosen from urn A was white given that exactly 2 white balls were selected is \(\boxed{\frac{7}{13}}\).