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13 Oct 2015, 11:27
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Peter took a 52-card deck of cards and removed all small cards, all spades, and all clubs.
The only remaining cards were 10, Jack, Queen, King, and Ace in the suits of heart and
diamond. Of these 10 cards, he deals himself 5 cards. What is the probability that Peter
deals himself at least one pair of cards?
A. 7/10
B. 17/33
C. 55/63
D. 66/165
E. 18/25
Note- This is a question from GMATPill exam. I do not have the answer to this question.
The only remaining cards were 10, Jack, Queen, King, and Ace in the suits of heart and
diamond. Of these 10 cards, he deals himself 5 cards. What is the probability that Peter
deals himself at least one pair of cards?
A. 7/10
B. 17/33
C. 55/63
D. 66/165
E. 18/25
Note- This is a question from GMATPill exam. I do not have the answer to this question.
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13 Oct 2015, 12:09
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gmatprep258
The only way that he doesn't deal a pair is if he deals 10,J,Q,K,A. So the probability of dealing a pair will be 1 - the probability of dealing 10,J,Q,K,A.
The total number of combinations is \(10*9*8*7*6\), which equals 30240
The number of ways to deal 10,J,Q,K,A or a variance of it is \(10*8*6*4*2\), which equals 3840
1 - 3840/30240 = 26400/30240. That will roughly equal 26/30, which when doubled is 52/60. That is very close to C, which is the answer. We can estimate on this problem because the answer choices are fairly far apart and because the other closest answer has a denominator of 25, which is not a factor of 30240.
C
13 Oct 2015, 14:49
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gmatprep258
P(at least 1 pair) = 1 - P(no Pairs)
P(no pairs) = number of hands dealt with no pairs / total possible number of hands
Number of hands dealt with no pairs has to be 10 J Q K A but with differing suits of hearts and diamonds. Since there are only two suits to pick from for a hand of 5 cards (just like a Coin Flip problem) total number of hands with no pairs = 2^5 = 32
Total possible number of hands is 10 C 5 = 10*9*8*7*6/ 5*4*3*2*1 = 252
Therefore 1-32/252 = 1- 8/63 = 55/63
Answer C
