mpevans wrote:
if 40 people get the chance to pick a card from a canister that contains 5 free passes to an amusement park mixed in with 35 blank cards what is the probability that the 40th person who picks will win?
I guess we have the situation when 40 people standing in a row and picking the cards one after another and check them in the end. We are asked what is the probability that 40th person win the pass?
If so, then probability of picking the pass will be the same for all 40 people - \(\frac{5}{40}\), (initial probability of picking the pass (\(\frac{5}{40}\)) will be the same for any person in a row).
AbhayPrasanna wrote:
35 - lose, 5 - win
Pick 5 people to win => 40C5 = total number of outcomes.
Favorable outcome is : First pick the 40th person, then pick any other 4.
=> 1*40C4
So, probability = 40C5 / 40C4
= 40!*36!*4! / (35!*5!*40!)
= 36/(35*5)
= 36 / 175
This way is also valid and must give the same result. The problem is that you calculated favorable outcomes incorrectly: when you pick 40th person to win, then you have only 39 left to pick 4 from, so # of favorable outcomes is \(1*C^4_{39}\). Also \(probability=\frac{# \ of \ favorable \ outcomes}{total \ # \ of \ outcomes}\) and you wrote vise-versa.
So \(P=\frac{1*C^4_{39}}{C^5_{40}}=\frac{36*37*38*39}{4!}*\frac{5!}{36*37*38*39*40}=\frac{5}{40}\).
Hope it helps.
Hi... thanks for the explanation Bunuel
But I do not understand how can the probability of selecting a free pass =5/40 for all in case we assume the people are picking the cards and keeping it with them.
won't it keep reducing as 4/39 for the second successfull fick of a free card...