Bag \( B_1 \) contains 6 white and 4 blue balls, Bag \( B_2 \) contains 4 white and 6 blue balls, and Bag \( B_3 \) contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If the ball is white, then the probability that the ball is drawn from Bag \( B_2 \) is:
Show Hint
- \( \frac{1}{3} \)
- \( \frac{2}{3} \)
- \( \frac{4}{15} \)
- \( \frac{2}{5} \)
The Correct Option is C
Approach Solution - 1
To solve this problem, we use Bayes' theorem to find the probability that the ball was drawn from Bag \( B_2 \), given that a white ball is drawn. Let's denote the events as follows:
- \( A_1 \): a ball is drawn from Bag \( B_1 \)
- \( A_2 \): a ball is drawn from Bag \( B_2 \)
- \( A_3 \): a ball is drawn from Bag \( B_3 \)
- \( W \): the ball drawn is white
We need to find \( P(A_2 \mid W) \), the probability that the ball is from Bag \( B_2 \) given that it is white. By Bayes' theorem:
\( P(A_2 \mid W) = \frac{P(W \mid A_2)P(A_2)}{P(W)} \)
Given that each bag is equally likely to be chosen, we have:
\( P(A_1) = P(A_2) = P(A_3) = \frac{1}{3} \)
Next, we calculate \( P(W \mid A_i) \), the probability of drawing a white ball from each bag:
\( P(W \mid A_1) = \frac{6}{10} = \frac{3}{5} \)
\( P(W \mid A_2) = \frac{4}{10} = \frac{2}{5} \)
\( P(W \mid A_3) = \frac{5}{10} = \frac{1}{2} \)
The total probability of drawing a white ball, \( P(W) \), is:
\( P(W) = P(W \mid A_1)P(A_1) + P(W \mid A_2)P(A_2) + P(W \mid A_3)P(A_3) \)
\( P(W) = \frac{3}{5}\cdot\frac{1}{3} + \frac{2}{5}\cdot\frac{1}{3} + \frac{1}{2}\cdot\frac{1}{3} \)
\( P(W) = \frac{1}{5} + \frac{2}{15} + \frac{1}{6} \)
To add these probabilities, we find a common denominator, which is 30:
- \( \frac{1}{5} = \frac{6}{30} \)
- \( \frac{2}{15} = \frac{4}{30} \)
- \( \frac{1}{6} = \frac{5}{30} \)
Thus:
\( P(W) = \frac{6}{30} + \frac{4}{30} + \frac{5}{30} = \frac{15}{30} = \frac{1}{2} \)
Using Bayes' theorem, we substitute back into \( P(A_2 \mid W) \):
\( P(A_2 \mid W) = \frac{\frac{2}{5}\cdot\frac{1}{3}}{\frac{1}{2}} \)
\( P(A_2 \mid W) = \frac{\frac{2}{15}}{\frac{1}{2}} = \frac{2}{15} \cdot 2 = \frac{4}{15} \)
Thus, the probability that the white ball was drawn from Bag \( B_2 \) is \( \frac{4}{15} \).
Approach Solution -2
There are three bags with the following distribution of balls:
Bag \( B_1 \): 6 white and 4 blue balls
Bag \( B_2 \): 4 white and 6 blue balls
Bag \( B_3 \): 5 white and 5 blue balls
One bag is chosen at random, and then one ball is drawn from it.
Step 2: Write down the probabilities.
The probability of selecting any one bag is equal since the choice is random:
\[ P(B_1) = P(B_2) = P(B_3) = \frac{1}{3} \]
The probability of drawing a white ball from each bag:
\[ P(W|B_1) = \frac{6}{10} = \frac{3}{5}, \quad P(W|B_2) = \frac{4}{10} = \frac{2}{5}, \quad P(W|B_3) = \frac{5}{10} = \frac{1}{2} \]
Step 3: Find the total probability of drawing a white ball.
Using the law of total probability:
\[ P(W) = P(B_1)P(W|B_1) + P(B_2)P(W|B_2) + P(B_3)P(W|B_3) \] \[ P(W) = \frac{1}{3}\left(\frac{3}{5} + \frac{2}{5} + \frac{1}{2}\right) \] \[ P(W) = \frac{1}{3}\left(\frac{3 + 2 + 2.5}{5}\right) = \frac{1}{3} \times \frac{7.5}{5} = \frac{1}{3} \times \frac{3}{2} = \frac{1}{2} \]
Step 4: Apply Bayes’ theorem.
We need to find the probability that the white ball was drawn from Bag \( B_2 \), i.e., \( P(B_2|W) \).
\[ P(B_2|W) = \frac{P(B_2)P(W|B_2)}{P(W)} \] \[ P(B_2|W) = \frac{\frac{1}{3} \times \frac{2}{5}}{\frac{1}{2}} = \frac{\frac{2}{15}}{\frac{1}{2}} = \frac{2}{15} \times 2 = \frac{4}{15} \]
Step 5: Conclusion.
The probability that the white ball was drawn from Bag \( B_2 \) is:
\[ \boxed{\frac{4}{15}} \]
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