Question

An archer hits a bullseye 64% of the time. What is the probability the archer hitsthe bullseye exactly 4 times during 10 total attempts?a. 242b. .365C..077d. 168

Answer 1

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:

`P(\bar{H})=1-P(H)=1-0.64=0.36`

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:

`P=P(H)^4\cdot P(\bar{H})^6=(0.64)^4\cdot(0.36)^6=0.16777\cdot0,0021767=0\text{.}0003652`

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:

`C(10,4)=(10!)/(4!(10-4)!)=(10!)/(4!6!)=(10\cdot9\cdot8\cdot7)/(4\cdot3\cdot2)=210`

So the final probability is:

`P^(\prime)=P\cdot C(10,4)=0.0003652\cdot210=0.077`

So the answer is C.