There are 15 apples in total. 4 apples go bad and the others are all fresh apples. Calculate the probability that: we picked the last bad apple at exactly the 9th pick.
Solution:
Assume that all of the apples are different.
Event Q: we picked the last bad apple at exactly the 9th pick.
P(Q) = #(3 bad apples are picked within first 8 picks and the 9th pick is also a bad apple) / sample space
sample space = #(possible order of 15 apples) = 15!
P(Q) = C(8, 3) * P(4, 4) * P(11, 11) / 15! = 8 / 195