5 A simple random sample of size five is selected from 52 ca

5. A simple random sample of size five is selected from 52 cards. Find the probability of each of the following events.

a. Hand contains all four 2\'s

b. Hand contains two aces and two sevens

c. hand contains three kings and the two other cards are of different denominations

Solution

a)

There are 52C5 ways to choose any 5 cards.

If we fix four cards to be 2\'s there are 48 ways to choose the last one.

Thus,

P(all four 2\'s) = 48/(52C5) = 0.0000184689 [ANSWER]

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b)

There are 4C2 ways to get two aces, 4C2 ways to get 2 7\'s, and 44 other ways to get the fifth card.

hence,

P = [(4C2)(4C2)(44)]/(52C5)

= 0.000609475 [ANSWER]

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c)

There are 4C3 ways to get 3 kings, and 48 ways to choose the one other card, and 44 ways to choose the last one.

thus,

P = [(4C3)*48*44]/(52C5)

= 0.003250531 [ANSWER]