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

Question

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

A. 7/10
B. 17/33
C. 55/63
D. 66/165
E. 18/25

chetan2u · Accepted Answer

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

prgmatbiz · Answer

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?

docc · Answer

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

The answer will be in the form    =

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.

nick1816 · Answer

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 · Answer

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

chetan2u · Answer

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}

Kinshook · Answer

Given: 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.

Asked: What is the probability that Peter deals himself at least one pair of cards?

Total cards = 5*2 = 10
Number of pairs = 5

Total number of ways to deal himself 5 cards = 10*9*8*7*6
Number of ways to deal himself no pair of cards = 10*8*6*4*2

The probability that Peter deals himself at least one pair of cards = 1 - The probability that Peter deals himself no pair of cards = 1 - (10*8*6*4*2/10*9*8*7*6) = 1 - 4*2/9*7 = 1 - 8/63 = 55/63

IMO C

reed990 · Answer

Anyone can help me to see why this is wrong?
Select 5 out of 10: 10C5: 252
Select 1 pair out of 5: 5C1: 5, and select 3 out of the rest 8 cards: 8C3: 56. Thus, for at least one pair, total number will be 5*56 = 280. I don't know why I got this larger than 252.

demawtf · Answer

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

A. 7/10
B. 17/33
C. 55/63
D. 66/165
E. 18/25

P (at least a pair) = 1- p(all different)

1) all different --> 5 cards, 1 10, 1 J, 1 Q, 1 K and 1 A. --> hearts and diamonds. So 2 choices for each value, 2^5=32 ways.
2) total ways 5 cards out of 10. 10!/5!5! =10*9*8*7*6=252
3) probability of all different 32/252=8/63
4) P(at least a pair) = 1-8/63=55/63 --> C

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

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