1. Basket A contains two white balls and three black balls, while Basket B contains a white ball and three black balls. Daniel randomly chooses one of the baskets and then randomly picks a ball from this basket. If he picked a white ball, what is the probability that his chosen basket was Basket A?
Answer: $\frac{8}{13}$
Solution: Let $P(A), P(B)$ be the probabilities of choosing Baskets $\mathrm{A}$ and $\mathrm{B}$, respectively, and let $P(W)$ be the probability of picking a white ball. The probability of picking a white ball from Basket A is $P(W \mid A)=2 / 5$ and the probability of picking a white ball from Basket $\mathrm{B}$ is $P(W \mid B)=1 / 4$. With $P(A)=P(B)=1 / 2$, we have $P(W)=P(A) P(W \mid A)+P(B) P(W \mid B)=(1 / 2)(2 / 5)+$ $(1 / 2)(1 / 4)=13 / 40$. Hence, the probability that a white ball Daniel picked came from the Basket $\mathrm{A}$ is
$$
P(A \mid W)=\frac{P(A) P(W \mid A)}{P(W)}=\frac{1 / 5}{13 / 40}=\frac{8}{13} .
$$