SOLUTION: There is a gambling game where equal-weight balls with numbers ranging from 1 to 15 are placed inside a tambiolo and mixed. Without replacement, three balls are chosen at random an

Question 1197439: There is a gambling game where equal-weight balls with numbers ranging from 1 to 15 are placed inside a tambiolo and mixed. Without replacement, three balls are chosen at random and are then chosen one by one.
(a) A certain casino customer named Mr. Han only has enough cash to place a total of 5 bets that evening. Mr. Han consistently bets that the 3 balls will be chosen, and that the greatest number will be at least as large as "12" among them. How likely is it that this will happen?
(b) What is the likelihood that Mr. Han will finally strike it lucky and win on the 5th bet after a run of bad luck?
Answer by Edwin McCravy(20054) (Show Source): You can put this solution on YOUR website!
Mr. Han will lose if the 3 balls are all chosen from balls 1 through 11.
That's 11C3 = 165 ways.
There are 15C3 = 455 ways the 3 balls can be selected, so there are 455-165=290
ways he can win.
So the probability that he wins a bet is = .
That's the answer to (a)
-----------------------------------
The probability that he loses a bet is . So the probability that he loses the first 4 and wins on the 5th is
which is about 0.011
Edwin

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