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Thread: Basketball player shootin E[X], V(X)?

  1. #1
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    Basketball player shootin E[X], V(X)?

    In a basketball shooting workout, a player keeps shooting until she makes 10 baskets. Suppose the probability that she makes any given shot is 0.7, and let X be the total number of shots she takes, calculate E[X], V(X)?

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    Re: Basketball player shootin E[X], V(X)?

    Expected Value and Variance for the Binomial Distribution:

    $$E[X] = 0.7X, V[X] = (0.7)(0.3)X$$
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    Re: Basketball player shootin E[X], V(X)?

    Thanks.
    but i think E[X] = 10/0.7 = 14
    and i am not sure about V(X)
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    Re: Basketball player shootin E[X], V(X)?

    Quote Originally Posted by Amanda2018 View Post
    Thanks.
    but i think E[X] = 10/0.7 = 14
    and i am not sure about V(X)
    Sorry, I misunderstood the original problem.

    Let's calculate variance a little differently. Since you have $E[X] = 14$, then

    $$V[X] = E[X^2]-14^2$$

    $$E[X^2] = (0.7)^{10}\sum_{n\ge 0}\dbinom{9+n}{n}(0.3)^n(10+n)^2 = \dfrac{10(11-0.7)}{0.7^2}$$

    So, you have:

    $$V[X] = \dfrac{696}{49}$$
    Thanks from Amanda2018
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    Re: Basketball player shootin E[X], V(X)?

    Quote Originally Posted by Amanda2018 View Post
    In a basketball shooting workout, a player keeps shooting until she makes 10 baskets. Suppose the probability that she makes any given shot is 0.7, and let X be the total number of shots she takes, calculate E[X], V(X)?
    Lets change the question.
    Toss a die repeatedly until a six appears the three times.
    Further, let $X$ be the number of the toss upon which the third six appears.
    Is it clear that $X\ge 3~?$ and $\mathcal{P}(X=3)=6^{-3}$. Moverover, $\mathcal{P}(X=9)=\dbinom{8}{2}\cdot 5^6\cdot 6^{-9}$

    In general, $n\ge 3\to \mathcal{P}(X=n)=\dbinom{n-1}{2}\cdot 5^{n-3}\cdot 6^{-n}$

    What are $\mathcal{E[X]}~\&~\mathcal{V[X]}~?$
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