Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls.

Question

Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked from bag B and mixed with the balls in bag A. Then a ball is randomly drawn from bag A. If the probability that the ball drawn is white is $\dfrac{p}{q}$, where $\gcd(p,q)=1$, then $p+q$ is equal to

Accepted Answer

Bag A: 9 white, 8 black (total 17). Bag B: 6 white, 4 black (total 10).

Case 1 (transferred ball white): $P(W_B)=\dfrac{6}{10}$. Bag A then has 10 white, 8 black (18 total): $P(W_A\mid W_B)=\dfrac{10}{18}$.

Case 2 (transferred ball black): $P(B_B)=\dfrac{4}{10}$. Bag A then has 9 white, 9 black (18 total): $P(W_A\mid B_B)=\dfrac{9}{18}$.

By total probability:

$$P(W_A) = \frac{6}{10}\cdot\frac{10}{18}+\frac{4}{10}\cdot\frac{9}{18} = \frac{60}{180}+\frac{36}{180}=\frac{96}{180}=\frac{8}{15}$$

So $p=8,\ q=15 \Rightarrow p+q=23$.

Solutions · 1