2.1.7 problem 9

(a)

$P\left(A_1\right|B)=\frac{P\left(A_1\right)P\left(B\right|A_1)}{P\left(B\right)}=\frac{P\left(A_1\right)}{P\left(B\right)}=\frac{P\left(A_2\right)}{P\left(B\right)}\frac{P\left(A_2\right)P\left(B\right|A_2)}{P\left(B\right)}=P(A_2\left|B\right)$.

(b)

If $B$ is implied by both $A_1$ and $A_2$, knowing that $B$ occured does not tip the probability of occurence in favor of either $A_1$ or $A_2$.

For example, let $A_1$ be the event that the card in my hand is the Ace of Spades. Let $A_2$ be the event that the card in my hand is the Ace of Hearts. Let $B$ be the event that there are $3$ aces left in the deck.

$B$ is implied by both $A_1$ and $A_2$, and $P\left(A_1\right)=P\left(A_2\right)$. Knowing that $B$ occured does not give one any information on whether they are holding the Ace of Spades or the Ace of Hearts, since $B$ would have occured in both cases. Thus, $P\left(A_1\right|B)=P(A_2\left|B\right)$.