A basketball player is historically an 82% free throw shooter. If she attempts 10 free throws, what is the probability she makes at least 8

Question

A basketball player is historically an 82% free throw shooter. If she attempts 10 free throws, what is the probability she makes at least 8 of them? (Round to three decimal places)

options for answers
1. 0.737
2. 0.820
3. 0.561
4. 0.296

please show how you figure it out that way i know for future reference.

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Valentina 1 week 2021-09-15T10:11:17+00:00 1 Answer 0

Answers ( )

    0
    2021-09-15T10:12:24+00:00

    Answer:

    1. 0.737

    Step-by-step explanation:

    For each free throw, there are only two possible outcomes. Either the player makes it, or he does not. The probability of the player making a free throw is independent of other free throws. So we use the binomial probability distribution to solve this question.

    Binomial probability distribution

    The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

    P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

    In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

    C_{n,x} = \frac{n!}{x!(n-x)!}

    And p is the probability of X happening.

    A basketball player is historically an 82% free throw shooter.

    This means that p = 0.82

    She attempts 10 free throws

    This means that n = 10

    What is the probability she makes at least 8 of them?

    P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10)

    P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

    P(X = 8) = C_{10,8}.(0.82)^{8}.(0.18)^{2} = 0.298

    P(X = 9) = C_{10,9}.(0.82)^{9}.(0.18)^{1} = 0.302

    P(X = 10) = C_{10,10}.(0.82)^{10}.(0.18)^{0} = 0.137

    P(X \geq 8) = P(X = 8) + P(X = 9) + P(X = 10) = 0.298 + 0.302 + 0.137 = 0.737

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