Explanation

Initially, bag A contains 9 white balls and 8 black balls.

So, total balls in bag A:

$$ 9 + 8 = 17 $$

Bag B contains 6 white balls and 4 black balls.

Total balls in bag B:

$$ 6 + 4 = 10 $$

One ball is randomly drawn from bag B.

Probability that the ball drawn from bag B is white:

$$ P(W_B)=\frac{6}{10}=\frac{3}{5} $$

Probability that the ball drawn from bag B is black:

$$ P(B_B)=\frac{4}{10}=\frac{2}{5} $$

Case 1: White ball is transferred from bag B to bag A.

Now, number of white balls in bag A becomes:

$$ 9 + 1 = 10 $$

Number of black balls remains:

$$ 8 $$

Total balls in bag A:

$$ 10 + 8 = 18 $$

Probability of drawing a white ball from bag A in this case:

$$ P(W_A|W_B)=\frac{10}{18}=\frac{5}{9} $$

Case 2: Black ball is transferred from bag B to bag A.

Now, number of white balls in bag A remains:

$$ 9 $$

Number of black balls becomes:

$$ 8 + 1 = 9 $$

Total balls in bag A:

$$ 9 + 9 = 18 $$

Probability of drawing a white ball from bag A in this case:

$$ P(W_A|B_B)=\frac{9}{18}=\frac{1}{2} $$

Using the law of total probability:

$$ P(W_A)=P(W_B)\cdot P(W_A|W_B)+P(B_B)\cdot P(W_A|B_B) $$

Substitute values:

$$ P(W_A)=\frac{3}{5}\cdot\frac{5}{9}+\frac{2}{5}\cdot\frac{1}{2} $$

$$ =\frac{3}{9}+\frac{1}{5} $$

$$ =\frac{5}{15}+\frac{3}{15} $$

$$ =\frac{8}{15} $$

Thus,

$$ \frac{p}{q}=\frac{8}{15} $$

So,

$$ p+q=8+15=23 $$

Hence, the correct answer is 23.

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