There are two bags $$B_1$$ and $$B_2$$. $$B_1$$ has two white and three black balls. $$B_2$$ has four white and two black balls. A ball is first drawn from $$B_1$$. Its colour is noted and it is put back into $$B_1$$. If the colour is white, then a ball is picked from $$B_2$$, else a ball is picked from $$B_1$$. What is the probability that the second ball is white?

To find the total probability that the second ball drawn is white, let us break down the problem into two distinct cases based on the outcome of the first draw from bag $$B_1$$.

Let us first define the composition of both bags:

• Bag $$B_1$$: 2 white and 3 black balls (Total = 5 balls)
  
• Bag $$B_2$$: 4 white and 2 black balls (Total = 6 balls)
  

A ball is drawn from $$B_1$$ first.

The probability of getting a white ball from $$B_1$$ is:

$$P(W_1) = \frac{2}{5}$$

Consequently, the probability of drawing a black ball from $$B_1$$ on the first try is:

$$P(B_1^{\text{black}}) = \frac{3}{5}$$

According to the given conditions, the process branches into two scenarios for the second draw:

Case 1: The first ball drawn is white

If the first ball is white, the second ball is drawn from bag $$B_2$$.

The probability of drawing a white ball from $$B_2$$ is:

$$P(W_2 \mid W_1) = \frac{4}{6} = \frac{2}{3}$$

The probability for this entire combined path is:

$$P_1 = P(W_1) \times P(W_2 \mid W_1) = \frac{2}{5} \times \frac{2}{3} = \frac{4}{15}$$

Case 2: The first ball drawn is black

If the first ball is not white, it is put back and a ball is picked again from bag $$B_1$$.

The probability of drawing a white ball from $$B_1$$ is:

$$P(W_2 \mid B_1^{\text{black}}) = \frac{2}{5}$$

The probability for this second combined path is:

$$P_2 = P(B_1^{\text{black}}) \times P(W_2 \mid B_1^{\text{black}}) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$$

Total Probability

Using the total probability theorem, we add the probabilities of both mutually exclusive cases:

$$P(\text{Second ball is white}) = P_1 + P_2 = \frac{4}{15} + \frac{6}{25}$$

Taking the least common multiple of the denominators (75):

$$P(\text{Second ball is white}) = \frac{20}{75} + \frac{18}{75} = \frac{38}{75}$$

The correct option is B.