Question

m 15 twice as likely to be probabilities that (a) C will win; (b) A will not win? 2. a new -, aver- , 0.34, reabil- 2.62. In
(a) two pairs (any two distinct face values occurring exactly twice); (b) four of a kind (four cards of equal face value). 2.
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Answer #1

a) 2 Pairs can be formed as = 13* 4C2 * 12* 4C2 * 44C1 / 52C5 =0.09507803

Hence we can say that 13 possibilities of getting 2 cards from a 4of a kind denomination

Now we have 12 remaining possibilities of selection 2 cards from a 4of a kind denomination

and now for the erst of teh denomination we have 44 cards from which we have 1 card to be selected

why 44 because 52-4(of 1st pair)-4(of 2nd pair)=44

b)for four of a kind we can select = 13* 4C4 * 48C1 / 52C5 =0.000240096

i.e. 4 cards of the same denomination 4C4 but we have 13 possible categories

So it's is 13 * 4C4

and for the rest we have 48 cards from which 1 card to be selected hence 48C1

While total possible selection are 52C5

Hope the above answer has helped you in understanding the problem. Please upvote the ans if it has really helped you. Good Luck!!

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