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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.

Solution

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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)=qk1pP(X=k)=q^{k-1}p

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

P(X=1)=qk1p=0.2011(0.80)=0.80P(X=1)=q^{k-1}p=0.20^{1-1}(0.80)=0.80

P(X=2)=qk1p=0.2021(0.80)=0.16P(X=2)=q^{k-1}p=0.20^{2-1}(0.80)=0.16

P(X=3)=qk1p=0.2031(0.80)=0.032P(X=3)=q^{k-1}p=0.20^{3-1}(0.80)=0.032

Add the corresponding probabilities:

P(X3)=P(X=1)+P(X=2)+P(X=3)=0.80+0.16+0.032=0.992=99.2%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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