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# A basketball player has made 80% of his foul shots during the season. Assuming the shots are independent, find the probability that in tonight's game he makes his first basket on one of his first 3 shots.

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The distribution of a variable that measures the number of trials until the first success is a Geometric distribution.

Definition geometric probability:

$P(X=k)=q^{k-1}p$

We also know $p=80\%=0.80$ (make shot) and $q=20\%=0.20$ (miss):

$P(X=1)=q^{k-1}p=0.20^{1-1}(0.80)=0.80$

$P(X=2)=q^{k-1}p=0.20^{2-1}(0.80)=0.16$

$P(X=3)=q^{k-1}p=0.20^{3-1}(0.80)=0.032$

$P(X\leq 3)=P(X=1)+P(X=2)+P(X=3)=0.80+0.16+0.032= 0.992=99.2\%$

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