The question is: A person is conducting a phone survey. Suppose that 5 of 6 people contacted will complete the survey. Define “success” as the event a person completes the survey and let Y be the number of failures before the third success. What is the probability that there are 10 failures before the third success?
Here is my thought process, n = 13, r = 3 and p = $5\over6$ or we can use the other formula and make y = 10, r = 3 and p should still be the same. Either way, I think p = $5\over6$ but the book is saying p = $1\over6$. Can anyone clarify why this is?
The answer is apparently:
$Pr(Y=10)=$$10+3-1\choose10$$({1\over6})^3({5\over6})^{10}$
I think it should be
$Pr(Y=10)=$$10+3-1\choose10$$({5\over6})^3({1\over6})^{10}$