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31 Mar 2022, 09:00
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C
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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
Are You Up For the Challenge: 700 Level Questions
A. 7/10
B. 17/33
C. 55/63
D. 66/165
E. 18/25
Are You Up For the Challenge: 700 Level Questions
Foreheadson
Joined: 22 Jun 2020
Last visit: 24 Sep 2022
Posts: 153
Given Kudos: 120
Location: Georgia
Concentration: Finance, General Management
Schools: Booth '25 (WL) Darden '25
GMAT 1: 720 Q51 V38
GPA: 3.71
31 Mar 2022, 11:57
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The probability that he gets no pair of cards = 1*8/9*6/8*4/7*2/6= 8/63
Than we need 1- that. 1-8/63=55/18
So the Answer is C.
Otherwise, using combinatorics:
total number of possibilities = 10C5=252
We need one of all the following 10,J,Q,K,A. each of these cards may have 2 colors. so 2^5=32
1-32/252=1-8/63=55/63.
Than we need 1- that. 1-8/63=55/18
So the Answer is C.
Otherwise, using combinatorics:
total number of possibilities = 10C5=252
We need one of all the following 10,J,Q,K,A. each of these cards may have 2 colors. so 2^5=32
1-32/252=1-8/63=55/63.
General Discussion
25 Dec 2023, 03:34
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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
Peter is choosing 5 cards from 10.
10 x 9 x 8 x 7 x 6/5! = 9 x 4 x 7
This will be the denominator in our answer.
A, D, and E have multiples of 5 in their denominators. 5 is not among the factors of our denominator.
B has a multiple of 11 in the denominator. 11 is not among the factors of our denominator.
The only possible answer is C.
--------------------------------------------------------------------------------------------------
Alternative approach:
9 x 4 x 7 = 252
The number of ways to select 5 cards without including any pairs is 10 x 8 x 6 x 4 x 2/5! = 32
The number of ways to select 5 cards with at least one pair is 252 - 32 = 220.
220/252 = 55/63
The correct answer is C.
A. 7/10
B. 17/33
C. 55/63
D. 66/165
E. 18/25
Peter is choosing 5 cards from 10.
10 x 9 x 8 x 7 x 6/5! = 9 x 4 x 7
This will be the denominator in our answer.
A, D, and E have multiples of 5 in their denominators. 5 is not among the factors of our denominator.
B has a multiple of 11 in the denominator. 11 is not among the factors of our denominator.
The only possible answer is C.
--------------------------------------------------------------------------------------------------
Alternative approach:
9 x 4 x 7 = 252
The number of ways to select 5 cards without including any pairs is 10 x 8 x 6 x 4 x 2/5! = 32
The number of ways to select 5 cards with at least one pair is 252 - 32 = 220.
220/252 = 55/63
The correct answer is C.
