Question 1197439
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 {{{290/455}}} = {{{58/91}}}. 
That's the answer to (a)

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The probability that he loses a bet is {{{1 - 58/91 = 33/91}}}.  So the probability that he loses the first 4 and wins on the 5th is 

{{{(33/91)^4(58/91)}}} which is about 0.011

Edwin</pre>