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Sagot :
If the probability of hitting the bullseye is 64% (P(H) = 0.64), then the probability of not hitting the bullseye (P(H_bar)) is:
[tex]P(\bar{H})=1-P(H)=1-0.64=0.36[/tex]Now, if we have 10 attempts, and the archer hits 4 times, then he misses 6 times.
So we have 4 cases of hitting (P(H)) and 6 cases of not hitting(P(H_bar)), and the probability is the product of the probabilities of each case:
[tex]P=P(H)^4\cdot P(\bar{H})^6=(0.64)^4\cdot(0.36)^6=0.16777\cdot0,0021767=0\text{.}0003652[/tex]We also need to multiply this probability by a combination of 10 choose 4, because the 4 hits among the 10 attempts can be any of the 10, in any order:
[tex]C(10,4)=\frac{10!}{4!(10-4)!}=\frac{10!}{4!6!}=\frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2}=210[/tex]So the final probability is:
[tex]P^{\prime}=P\cdot C(10,4)=0.0003652\cdot210=0.077[/tex]So the answer is C.
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