Question

Bag I contains 4 white and 6 black balls.
Bag II contains 4 white and 3 black balls.
One ball is drawn at random from any one of the two bags and it is found to be a black ball. The probability that the black ball was drawn from Bag I is ______ (rounded off to two decimal places).

Hint:

Use Bayes' theorem: weight each bag's black-ball chance by \(\frac{1}{2}\) and take Bag I's share of the total.

Answer 1

Correct Answer: 0.58

Step 1: Define events and set up Bayes' theorem.
Let \(E_1\): Bag I is chosen, \(E_2\): Bag II is chosen, and \(B\): a black ball is drawn. Since a bag is picked at random from the two available bags, \(P(E_1) = P(E_2) = \frac{1}{2}\). We need \(P(E_1 \mid B)\), the probability that Bag I was the source, given that the ball drawn turned out to be black. By Bayes' theorem:
\[ P(E_1 \mid B) = \frac{P(E_1)\,P(B \mid E_1)}{P(E_1)\,P(B \mid E_1) + P(E_2)\,P(B \mid E_2)} \]

Step 2: Find the conditional probabilities of drawing black from each bag.
Bag I has 4 white and 6 black balls, 10 balls in total, so:
\[ P(B \mid E_1) = \frac{6}{10} = 0.6 \]
Bag II has 4 white and 3 black balls, 7 balls in total, so:
\[ P(B \mid E_2) = \frac{3}{7} \approx 0.42857 \]

Step 3: Compute the total (marginal) probability of drawing a black ball.
\[ P(B) = P(E_1)P(B\mid E_1) + P(E_2)P(B\mid E_2) = \frac{1}{2}(0.6) + \frac{1}{2}(0.42857) \]
\[ P(B) = 0.3 + 0.214286 = 0.514286 \]

Step 4: Apply Bayes' theorem to get \(P(E_1 \mid B)\).
\[ P(E_1 \mid B) = \frac{0.3}{0.514286} = 0.5833 \]
Rounding to two decimal places:
\[ P(E_1 \mid B) \approx 0.58 \]

Step 5: Final Answer.
The probability that the black ball was drawn from Bag I is approximately 0.58.
\[ \boxed{0.58} \]