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# Playing card probability

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A bridge hand consists of 13 cards dealt at random from an ordinary deck of 52 playing cards.
a)How many possible bridge hands are there?
b) Find the probability of being dealt a bridge hand that contains exactly two of four aces.
c) Find the probability of being dealt an 8-4-1 distribution, that is, eight cards of one suit, four of another, and one of another.
d) Determine the probability of being dealt a 5-5-2-1 distribution.
e) Determine the probability of being dealt a hand void in a specific suit.

https://brainmass.com/statistics/probability/playing-card-probability-13956

#### Solution Preview

a.)
<br>Possible bridge hands (S)= 52C13 = 52!/(13!*39!) --Answer
<br>
<br>b.)
<br>number of cases in which there are exactly two aces in a hand (n)
<br>=> n =4C2 * 48C11 = (4!/(2!*2!))*(48!/(11!*37!))
<br>=> n =(4*3/2)*(48!/(11!*37!))
<br>=> n = 6*(48!/(11!*37!))
<br>hence, the probability of exactly two aces in a hand:
<br>p = n/S = 6*(48!/(11!*37!))/(52!/(13!*39!))
<br>=> p = 6*48!*13!*39/(52!*11!*37!)
<br>=> p = 6*13*12*39*38/(52*51*50*49)
<br>=> 3*6*13*19/(17*25*49) = 4446/20825= 0.21349 --Answer
<br>
<br>c.)
<br>no. of ways of for getting 8-4-1 distribution
<br>n = 4C1*13C8 * 3C1*13C4 * 2C1*13C1
<br>=> n = 4*(13!/(8!*5!)) * 3*(13!/(4!*9!)) * 2*(13!/(1!*12!))
<br>=> n = ...

#### Solution Summary

A bridge hand consists of 13 cards dealt at random from an ordinary deck of 52 playing cards.
a)How many possible bridge hands are there?
b) Find the probability of being dealt a bridge hand that contains exactly two of four aces.
c) Find the probability of being dealt an 8-4-1 distribution, that is, eight cards of one suit, four of another, and one of another.
d) Determine the probability of being dealt a 5-5-2-1 distribution.
e) Determine the probability of being dealt a hand void in a specific suit.

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