Probability Part 3

Concept:Use the law of total probability to combine the two possible transfer events from bag B to bag A.Explanation:The ball transferred from bag B can be white or black.Case 1: A white ball is transferred.Probability of picking white from bag B = $\frac{6}{10}$.After transfer, bag A contains 10 white and 8 black balls (total 18).Probability of drawing white from bag A = $\frac{10}{18}$.Case 2: A black ball is transferred.Probability of picking black from bag B = $\frac{4}{10}$.After transfer, bag A contains 9 white and 9 black balls (total 18).Probability of drawing white from bag A = $\frac{9}{18}$.Using total probability:$P(\text{white from A}) = \frac{6}{10} \times \frac{10}{18} + \frac{4}{10} \times \frac{9}{18}$.$= \frac{60}{180} + \frac{36}{180} = \frac{96}{180} = \frac{8}{15}$.Given $P = \frac{p}{q}$ with $\gcd(p,q)=1$, so $p=8$, $q=15$.Thus $p+q = 8+15 = 23$.Answer:$23$ (Option C).