I am really stuck on this question on my last homework and did not have time to stop by office hours. If anyone can walk me through this, I would really appreciate it.
A marksman scores a bull's eye on 90% of his shots.
a) What is the probability that he gets at least eight bull's eyes if he shoots ten times?
b) If he shoots until he gets eight bull's eyes, what is the probability that he needs at most ten shots?
For part a) I did ${10\choose{8}}(.9)^8(.1)^2$. When worked through, I get .19371. I know the answer is .92981 for both questions but cannot figure out how to get there. Thank you in advance!
If $X$ is the random variable corresponding to the number of successful shots, then $X \sim B(0.9,10)$.
For (a), we want to compute $P(X\ge 8)$ and can be done as follow, \begin{align} P(X\ge8) &= P(X=8) + P(X=9) + P(X=10) \\ &= \binom{10}{8}(.9)^8(.1)^2 \binom{10}{9}(.9)^9(0.1) + \binom{10}{10}(0.9)^{10} \\ &= 0.9298092... \end{align}