Maybe sounds weird, but this actually applies to programming:
Say you have 100 tickets in a raquet and there are 3 which entitles to a prize. If you buy 10, what is the probability that at least one is a winner?
+1
A:
The probability that any randomly drawn ticket wins is variable, since the number of tickets remaining declines as you draw them. Let's say you draw the 3 winners right in a row. At the time of drawing, their probabilities were 3/100, then 2/99, then 1/98 (after each successive winner gets drawn). But you probably won't draw 3 in a row, so you well could end up having drawn 50 tickets and still have 1 winner left, leaving you with 1/50 probability at that point. Etc.
You could do a Monte Carlo simulation. In that case, this would actually be an appropriate programming question!
Or you could say "3 out of 100 is close enough." If you choose the latter, then the answer is:
Probability that none of the 10 tickets is a winner (97 out of 100 won't win and there are 10 pseudo-independent draws): (0.97)^10 = 73.7%
Probability that at least one of the 10 tickets is a winner: 1 - (0.97)^10 = 26.3%
Edit: See the answer by Roger Wilcocks for a more accurate estimate that is better than (0.97)^10.
Edit2: Actually, Roger's answer is incorrect.
jpp
2010-07-07 05:08:10
You forgot to allow that as you draw non-winning tickets, you improve the odds of the next tixket being a winner.
Roger Willcocks
2010-07-07 05:09:46
Yep, I went back to edit it and then saw your post, so credit to you.
jpp
2010-07-07 05:11:08
Wow! By that logic, if I buy 40 tickets, I am guaranteed to win! :-)
Franci Penov
2010-07-07 05:21:40
Actually, Roger's answer is even less correct than mine. The probability of not getting any winning tickets:(97/100) * (96/99) * (95/98) ... but each of those is closer to ~97/100 than they are to his answer.
jpp
2010-07-07 05:28:58
My approach gives 0.737 as the probability you get no winners. His gives 0.456 as the probability that you get no winners. And the true answer is (97/100)*(96/99)...etc which is 0.727. So mine, though not quite correct, is far closer to the truth than his.
jpp
2010-07-07 05:31:30
@jpp you are right. I should have done the full factorial evaluation :)
Roger Willcocks
2010-07-07 05:33:03
A:
Oh looky, a factorial question.
1 - the odds you didn't win any.
so,
the first ticket 0.97 *
the second ticket 0.96 *
the third ticket 0.95 *
. . .
Gives you the odds you didn't win at all.
Roger Willcocks
2010-07-07 05:08:33