Let $A$ be the event of a correct choice among $n$ alternatives. Let $B$ be the event of knowing prior to the selection which of the alternatives are correct. Suppose $P(B)=p, 0 < p < 1$ (a) Determine $P(B|A)$ (b) Show $P(B|A) \ge P(B)$ and show $P(B|A)$ increases with increasing $n$ for fixed $p$.
This is Paul E. Pfeiffer Concepts of Probability Theory problem 2-25 which I am working exercises from.
Here is my solution - I would love corrections or suggestions for better approaches.
(a)
We assume that if the answer is know prior to selection that the correct choice is always made so that $P(A|B) = 1$ and by definition, $P(A|B) = {P(AB) \over P(B)}$ so that $P(AB) = P(B) = p$
We assume if the answer is not known prior to selection that every alternative is equally likely to be selected (ie some choices are not more attractive as guesses) so that $P(A|B^c) = {1 \over n} = {P(AB^c) \over P(B^c)} = {P(AB^c) \over 1-p}$ and thus $P(AB^c) = {1-p \over n }$
Thus $P(A) = P(AB) + P(AB^c) = p + {1 - p \over n}$
Then $P(B|A) = {P(AB) \over P(A)} = {p \over p + {1-p \over n}}$
(b)
$P(B|A) = {P(AB) \over P(A)} = {P(B) \over P(A)} \gt P(B)$ since $0 \lt P(A) \le 1$
Consider $P(B|A) = {p \over p + {1-p \over n}}$
Since $0 \lt 1-p \lt 1$, we see that the denominator decreases as $n$ increases. If the denominator decreases, for a fixed numerator, the overall quantity increases. Thus as $n$ increases, $P(B|A)$ increases.
Looks good.
$$P(B|A)=\frac{P(B)P(A|B)}{P(A)}=\frac{P(B)P(A|B)}{P(B)P(A|B) + P(B^C)P(A|B^C)}=\frac{p\cdot 1}{p\cdot 1 +(1-p)\frac1n }=\frac{p}{p+\frac{1-p}n}$$