Question 1120908
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(1) # of ways of choosing 2 of the 13 denominations: 13C2 = (13*12)/2 = 78<br>
(2) # of ways of choosing 1 of those 2 denominations to be the one to have 3 of a kind: 2C1 = 2<br>
(3) # of ways of choosing 3 of the 4 cards of the first denomination: 4C3 = 4<br>
(4) # of ways of choosing 2 of the 4 cards of the second denomination: 4C2 = 6<br>
(5) # of ways of choosing 3 cards of one denomination and 2 of another: 78*2*4*6 = 3744<br>
Probability of choosing 3 cards of one denomination and 2 of another: 3744/(52C5) = 3744/2598960 = .00144  (to 5 decimal places)