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?

Question

Gaven Francis

Mathematics

11 June, 03:12

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?

+5

Answers (1)

Know the Answer?

Cecilia Mays · Answer 1

89.44%.

Step-by-step 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).