06-05-2005, 05:33 PM
Quote:
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Originally Posted by ankursanghi
khanna sahib,
could you please explain, for a probability-phobic guy, how u achieved the answer??
rgds,
ankur
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after removing two cards which are same( lets say two 1s)
we are left with 36 cards ( 2,3,4,5,6,7,8,9,10 ) i.e. 4 cards of each type and two cards numbered 1......now if a draw 2 cards and want those two cards to be of same type then either both are 1s or both 2s and so on.......out of two 1s, we can select two 1s in C(2,2) ways....out of four 2s we can select two 2s in C(4,2) ways...similarly number of ways of selecting cards numbered (3-10) are C(4,2) each.....hence total number of ways of selecting 2 same cards from these 38 cards are C(4,2)+C(4,2).....9 times + C(2,2) = 9*C(4,2)+C(2,2)
..total number of ways of selcting 2 cards from a set of 38 cards are C(38,2)...
so prob = favourable cases/total cases =
{9*C(4,2)+C(2,2)}/C(38,2)
rgds