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22 December, 13:42

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?

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Answers (1)
  1. 22 December, 15:16
    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).
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