Let's say we have drawn five-card hands A, B and C (without replacement). This means that there are no duplication of cards in each hand.
The number of ways of choosing the exact 15 cards from a 52-card deck is D1=C(52,15), where
C(n,r)=n!/(r!(n-r)!)
The number of ways the 15 cards can be arranged into the three given decks A,B,C is
D2=C(15,5)*C(10,5)*C(5,5)/3!
We divide by 3! because the order of hands does not count, and there are 6 ways the hands can be formed (ABC,ACB,BAC,BCA,CAB,CBA).
Thus the probability of drawing the given 3 five-card hands is
1/(D1*D2)
=1/(2598960*126126)
=3.05×10-12
Can someone please explain this to be, so that I can understand this problem?
Select three different five-card combinations or five-card hands from your favorite card game that utilizes a standard 52-card deck containing four suits (clubs, hearts, diamonds, and spades), with each suit containing 13 cards with numbers 2–10 and face cards ace, king, queen, and jack. Then, do the following:
Using the concept of dependent probabilities, determine the odds that you would draw these hands (card combinations) directly from a deck of cards.
Determine the probability that you would not draw these hands (card combinations) directly from a deck of cards.
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