Question 873634
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Hi
5 cards are drawn, without replacement
Think of choosing 5 Cards of a Particular Kind w/o Replacement: Particular/ALL
P(all diamonds)= {{{(13C5)/(52C5) }}} = {{{(13*12*11*10*9)/(52*51*50*49*48)}}}
P(all Aces)= 0 (only 4 Aces)
P(all Aces and one other) = {{{(4C4)(39C1)/(52C5)}}}
P(Choosing 4C and 1 nonC) = {{{(13C4)(39C1)/(52C5)}}}