A professional bowler works on his bowling game so that he bowls over 200 80% of the time. What is the probability that he exceeds 200 in at least 9 of his next 10 games?

Question

A professional bowler works on his bowling game so that he bowls over 200 80% of the time.  What is the probability that he exceeds 200 in at least 9 of his next 10 games?

anonymous · Answer

Are you familiar with the binomial distribution?

andu1854 · Answer

Its been a few years since I have taken Stats and relearning it as I go, but I couldn't figure out how the answer is .3758.  I tried setting up the problem as 10 C 9 * .80^9 *.26^1, but not sure if this was the correct way to set it up or if I need to do one more step since you are looking to score better than what you normally score...

anonymous · Answer

Okay, you're on the right track; but first off, the first part should be $\large _{10}C_9 (.80)^9(1-.80)^1 = _{10}C_9 (.80)^9(.\color{red}{20})^1$.  Now, the key words that you want to pay close attention to in the question statement "What is the probability that he exceeds 200 in at least 9 of his next 10 games?" are *at least*.

In the context of this problem, "at least 9 of the next 10 games" means that you consider the probability of him accomplishing the feat in exactly 9 of 10 games (which you already have done), and then add this result to the the probability that he accomplishes the feat in all 10 games, which in this case would be given by $\large _{10}C_{10}(.80)^{10}(.20)^{0}$.

Therefore, the probability that he exceeds 200 in at least 9 of the next 10 games is  $\large _{10}C_{9}(.80)^{9}(.20)^{1} + _{10}C_{10}(.80)^{10}(.20)^{0}=\ldots$.

Does this clarify things a little bit?

andu1854 · Answer

yes thank you very much