Ask Question
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)
  1. 11 June, 03:17
    0
    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).
Know the Answer?
Not Sure About the Answer?
Find an answer to your 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 ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers