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1025867@etusd.org

 
 
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1025867@etusd.org
by shawn luo - Wednesday, February 12, 2020, 4:14 PM
 

for 8.27, why do you multiply 13 choose 1 and 12 choose 3.

 
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Re: 1025867@etusd.org
by Areteem Professor - Thursday, February 13, 2020, 10:33 AM
 

We want to find the probability of being dealt exactly one pair when we are dealt  5 cards.

The $\displaystyle \binom{13}{1}$ chooses the rank of the pair, and the $\displaystyle\binom{12}{3}$ chooses three other distinct ranks so that we have one and only one pair. Then $\displaystyle \binom{4}{2}$ chooses which two suits will be chosen for the pair, and $\displaystyle \binom{4}{1}^3$ chooses the suit for each of the single cards.

Thus, the chances of getting exactly one pair are $$\dfrac{\displaystyle \binom{13}{1}\binom{12}{3}\binom{4}{2}\binom{4}{1}^3}{\displaystyle\binom{52}{5}}.$$