Trouble with homework again. Don’t need the answer, just a heads up on what I’m doing wrong:
"A basketball coach ends every practice with a free throw drill in which he requires each player to sink 13 free throws. Suppose one player has free throw percentage 38 %.
(a) What is the probability that it takes him exactly 45 tries to complete the drill? "
Here I did (45 choose 13)* (.38^13) * (.62^32). This comes out to about .057, which seems like a reasonable number but is incorrect. Teacher did pretty similar problem in class and that appears to be the formula he used for it.
I think the problem is that your method includes, for example, the case where the player first succeeds thirteen times in a row, and then fails 32 times. The probability you want is the probability of the player hitting 12 out of the first 44, and ten hitting again on the 45th throw.
I think Sofis is correct. The probability you found is the probability of getting exactly 13 successes in the first 45 shots. This can happen if the 13th success occurs on the 45th shot, or if it occurs earlier. If it occurs earlier, then it doesn’t take 45 shots to complete the drill, so this case shouldn’t be included in calculating the probability. Instead, find the probability of getting exactly 12 successes in 44 trials, and a success on the 45th, like Sofis said.
Alright, I see what you’re saying now, but I’m clueless as to how to do it. The probability of making the 45th shot is just .38, which if adding that to the 44 choose 12 etc… would clearly result in too high a number. Then I tried 45 choose 1 * .38 * .62^44, but that number is too small to affect the answer (it only verifies the first 4 sig figs).
Consider two independent events. The first is the event that exactly 12 successes occur in the first 44 trials. The second event is that the 45th shot is a success. What you want is the probability that both of these events occur. Since they are independent, this is just the product of the probabilities for each event.