Question

When shooting two consecutive free throws

during the regular season, a basketball player
makes the first free throw 78% of the time.
If he makes the first free throw, he makes
the second one 88% of the time, but he only
makes the second free throw 52% of the
time after missing the first one.
When he shoots a pair of free throws
in the team’s first playoff game, what
is the probability that he makes at least
one free throw?

Answer 1

89.44%.

Explanation:

Let's work out the probability he misses both throws:

Prob( he misses both throws) = (1-0.78) * ( 1 - 0.52)

= 0.22*0.48

= 0.1056.

So the probability he makes at least one free throw = 1 - 0.1056

= 0.8944.

(It is 1 - 0.1056 because the default of missing both throws is either making one throw on first or second attempt, or making both throws).