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# Peter took a 52-card deck of cards and removed all small cards, all sp

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Math Expert
Joined: 02 Sep 2009
Posts: 58402
Peter took a 52-card deck of cards and removed all small cards, all sp  [#permalink]

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16 Aug 2018, 08:36
11
00:00

Difficulty:

65% (hard)

Question Stats:

61% (03:26) correct 39% (03:23) wrong based on 41 sessions

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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

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Math Expert
Joined: 02 Aug 2009
Posts: 7971
Re: Peter took a 52-card deck of cards and removed all small cards, all sp  [#permalink]

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16 Aug 2018, 09:08
1
Bunuel wrote:
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

Best way is to do 1-P(all different)
Ways all are different = 10*8*6*4*2
Ways to pick 5 out of 10 = 10*9*8*7*6
P(all different) = $$\frac{10*8*6*4*2}{10*9*8*7*6}=\frac{8}{63}$$
Ans = 1-8/63=55/63

C
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Joined: 12 Aug 2019
Posts: 8
Re: Peter took a 52-card deck of cards and removed all small cards, all sp  [#permalink]

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16 Oct 2019, 13:18
1
chetan2u wrote:
Bunuel wrote:
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

Best way is to do 1-P(all different)
Ways all are different = 10*8*6*4*2
Ways to pick 5 out of 10 = 10*9*8*7*6
P(all different) = $$\frac{10*8*6*4*2}{10*9*8*7*6}=\frac{8}{63}$$
Ans = 1-8/63=55/63

C

chetan2u Bunuel

I did this :

answer = 1-P(all 5 hearts OR all 5 diamonds)
= 1 - (1/252 +1/252)
= 125/126

can you explain the error in my method?
Intern
Joined: 15 Oct 2019
Posts: 1
Re: Peter took a 52-card deck of cards and removed all small cards, all sp  [#permalink]

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16 Oct 2019, 13:55
Bunuel wrote:
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

I started by calculating the total number of potential hands. 10 choose 5 = 252.

The answer will be in the form [Number of hands with at least 1 pair][/Total number of hands] = [N][/252]

None of the answers have a denominator of 252 so the actual denominator must be a factor of 252.
By creating a factor tree I found the prime factors of 252 are 2*2*3*3*7

Looking at the choices:
A) Has a denominator of 10. 5 is a prime factor of 10 but not 252 so this answer is INCORRECT.
B) 33 has a prime factor of 11. INCORRECT.
C) 63's prime factors are 3*3*7. POSSIBLE answer
D) 165 has a prime factor of 5. INCORRECT.
E) 25 has a prime factor of 5. INCORRECT.

By process of elimination, I found answer C to be the only possible answer.
Director
Joined: 19 Oct 2018
Posts: 973
Location: India
Re: Peter took a 52-card deck of cards and removed all small cards, all sp  [#permalink]

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16 Oct 2019, 14:06
You missed 30 cases.

When he had 4 diamond cards and 1 heart card= 5!/4!=5
When he had 3 diamond cards and 2 heart cards= 5!/3!2!= 10
When he had 2 diamond cards and 3 heart cards= 5!/2!3!= 10
When he had 1 diamond card and 4 heart cards = 5!/4!=5

prgmatbiz wrote:
chetan2u wrote:
Bunuel wrote:
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

Best way is to do 1-P(all different)
Ways all are different = 10*8*6*4*2
Ways to pick 5 out of 10 = 10*9*8*7*6
P(all different) = $$\frac{10*8*6*4*2}{10*9*8*7*6}=\frac{8}{63}$$
Ans = 1-8/63=55/63

C

chetan2u Bunuel

I did this :

answer = 1-P(all 5 hearts OR all 5 diamonds)
= 1 - (1/252 +1/252)
= 125/126

can you explain the error in my method?
Intern
Joined: 12 Aug 2019
Posts: 8
Re: Peter took a 52-card deck of cards and removed all small cards, all sp  [#permalink]

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16 Oct 2019, 18:57
looks like i really misunderstood the question... they were asking for pairs not shapes...

so prob to get all different say Jacks = 2/10 * 8/9 * 6/8 * 4/7 * 2/6 (2/10 bcos we have 2 jacks out of 10, 8/9 bcos we want anything other than jack from the remaining nine cards and so on....) = 8/315

prob to get all different = 5 * 8/315 = 8/63 (multiply by 5 as we have 5 possibilities)

answer to the question = 1 - prob all different = 1 - 8/63 = 55/63

hope this is correct

nick1816 wrote:
You missed 30 cases.

When he had 4 diamond cards and 1 heart card= 5!/4!=5
When he had 3 diamond cards and 2 heart cards= 5!/3!2!= 10
When he had 2 diamond cards and 3 heart cards= 5!/2!3!= 10
When he had 1 diamond card and 4 heart cards = 5!/4!=5

prgmatbiz wrote:
chetan2u wrote:

chetan2u Bunuel

I did this :

answer = 1-P(all 5 hearts OR all 5 diamonds)
= 1 - (1/252 +1/252)
= 125/126

can you explain the error in my method?
Math Expert
Joined: 02 Aug 2009
Posts: 7971
Re: Peter took a 52-card deck of cards and removed all small cards, all sp  [#permalink]

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16 Oct 2019, 19:07
1
prgmatbiz wrote:
looks like i really misunderstood the question... they were asking for pairs not shapes...

so prob to get all different say Jacks = 2/10 * 8/9 * 6/8 * 4/7 * 2/6 (2/10 bcos we have 2 jacks out of 10, 8/9 bcos we want anything other than jack from the remaining nine cards and so on....) = 8/315

prob to get all different = 5 * 8/315 = 8/63 (multiply by 5 as we have 5 possibilities)

answer to the question = 1 - prob all different = 1 - 8/63 = 55/63

hope this is correct

prgmatbiz wrote:
chetan2u wrote:

chetan2u Bunuel

I did this :

answer = 1-P(all 5 hearts OR all 5 diamonds)
= 1 - (1/252 +1/252)
= 125/126

can you explain the error in my method?

Yes, but a better and straight forward way would be..
First could be any of the card, so 10/10..
Next could be anything of remaining 9 except the pair of the first one, so (9-1)/9..and so on..
prob that all are different = $$\frac{10}{10}*\frac{8}{9}*\frac{6}{8}*\frac{4}{7}*\frac{2}{6}=\frac{8}{63}$$
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Re: Peter took a 52-card deck of cards and removed all small cards, all sp   [#permalink] 16 Oct 2019, 19:07
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