Bag B1 contains 6 white and 4 blue balls, Bag B2 contains 4 white and 6 blue balls, and Bag B3 contains 5 white and 5 bl

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Bag B1 contains 6 white and 4 blue balls, Bag B2 contains 4 white and 6 blue balls, and Bag B3 contains 5 white and 5 blue balls. One of the bags is selected at random and a ball is drawn from it. If…

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Given: Three bags are available, and one bag is chosen at random. The compositions are: $B_1$ has $6$ white and $4$ blue balls, $B_2$ has $4$ white and $6$ blue balls, and $B_3$ has $5$ white and $5$ blue balls.

Find: The conditional probability that the white ball was drawn from Bag $B_2$ .

Let

A_1 = \text{ball drawn from Bag } B_1, \quad A_2 = \text{ball drawn from Bag } B_2, \quad A_3 = \text{ball drawn from Bag } B_3, \quad W = \text{white ball drawn}

We need to find $P(A_2 \mid W)$ .

Since each bag is equally likely to be selected,

P(A_1) = P(A_2) = P(A_3) = \frac{1}{3}

The probability of drawing a white ball from each bag is

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}

Using the law of total probability,

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)

So,

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}

Taking the common denominator $30$ ,

\frac{1}{5} = \frac{6}{30}, \quad \frac{2}{15} = \frac{4}{30}, \quad \frac{1}{6} = \frac{5}{30}

Therefore,

P(W) = \frac{6}{30} + \frac{4}{30} + \frac{5}{30} = \frac{15}{30} = \frac{1}{2}

Now apply Bayes' theorem:

P(A_2 \mid W) = \frac{P(W \mid A_2)P(A_2)}{P(W)}

Substituting the values,

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}

Therefore, the probability that the white ball is drawn from Bag $B_2$ is $\frac{4}{15}$ . The correct option is C.

Given: A bag is chosen uniformly from B1, B2, B3 and then one ball is drawn. A white ball is observed. Find: The probability that the chosen bag was B2. Use Bayes' theorem for conditional probability: P(B2 W) = P(W B2)P(B2)P(W) First compute the needed probabilities: P(B1) = P(B2) = P(B3) = 13 P(W B1) = 610 = 35, P(W B2) = 410 = 25, P(W B3) = 510 = 12 Now find the total probability of getting a white ball: P(W) = P(B1)P(W B1) + P(B2)P(W B2) + P(B3)P(W B3) P(W) = 13 35 + 13 25 + 13 12 P(W) = 15 + 215 + 16 = 6+4+530 = 1530 = 12 Finally, P(B2 W) = 25 13 12 = 215 2 = 415 Hence, the required probability is 415.

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