Question:
"Suppose that a basketball player can make a free throw 60% percent of the time. Let X equal the minimum number of free throws that this player must attempt to make a total of 10 shots. Find P(X = 16)."
Thank you so much in advance. I really need help with this one. I couldn't understand why it isn't simply $ (15 C 9)(.6^9)(.4^6)$ Aside from putting it directly into the equation, I tried to think of it combinatorically where I laid out 16 slots, knowing that at least 10 have to be successes, but since we are observing where X=16 the last slot is a guaranteed hit and therefore just 1 times the remaining amount of hits and misses. What's wrong with my idea? Please help, thank you!
Best Answer
We have $X=16$ if her $10$-th success occurred on the $16$-th free throw attempt.
We assume that the events "success" on the various throws are independent.
The $10$-th success came on the $16$-th attempt precisely if the following happened: (i) She had exactly $9$ successes on the first $15$ trials and (ii) her $16$-th free throw resulted in a success.
The probability of (i) is $\binom{15}{9}(0.6)^{9}(0.4)^{6}$.
Given that (i) happened, the probability of success on the $16$-th attempt is $0.6$.
It follows that the probability that (i) and (ii) happened is $$\binom{15}{9}(0.6)^9(0.4)^{6}(0.6).$$
Remark: Your calculation was almost right. You had calculated the probability of $9$ successes in the first $15$ trials. But we must in addition get a success on the $16$-th trial.