Question

When at the free-throw line for two shots, a basketball player makes at least one free throw 90% of the time. 80% of the time, the player makes the first shot, 80% of the time the player makes the second shot and 70% of the time she makes both shots. What is the conditional probability that the player makes the second shot given that she missed the first? Express your answer as a percentage to the nearest percentage point.

Answer 1

Let
\(F\) = player makes the first shot
\(\overline{F}\) = player does not make the first shot
\(S\) = player makes the second shot

You are given:
\( P(F \cup S)=0.9\) since at least one shot means they got First shot OR Second short or both (and there are only 2 shots in total)
\(P(F)=0.8\)
\(P(S)=0.8\)
\(P(F \cap S)=0.7\)

You are asked:
to find \(P(S|\overline{F})\)

\[ P(S|\overline{F})=\frac{P(S\cap\overline{F})}{P(\overline{F})}\]
Notice that
\(P(S)=P(S\cap F)+P(S\cap\overline{F})\\
\implies P(S\cap \overline{F})=P(S)-P(S\cap F)=0.8-0.7=0.1\)

And
\(P(\overline{F})=1-P(F)=1-0.8=0.2\)

So:
\[ P(S|\overline{F})=\frac{P(S\cap\overline{F})}{P(\overline{F})}=\frac{0.1}{0.2}=0.5=50\%\]

Answer 2

Thank you very much.