Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
(A) 8/33
(B) 62/165
(C) 17/33
(D) 103/165
(E) 25/33
Can someone please tell me the approach to do this question?
Probability - playing cards
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Hi,
Interesting problem ... Perhaps amongst the tough ones in GMAT.
Atleast 1 card repeating = total - No cards repeating.
Total card possibilities = 12 C 4 = 495
No card repeatings = 5 ways of achieving this
4 cards from deck1 (or)
3 cards from d1 and 1 of the other cards from d2 (or)
2 cards from d1 and 2 of the other cards from d2 (or)
1 card from d1 and 3 of the other cards from d2 (or)
4 cards from d2
=
6C4 +
6C3 times 3C1 +
6C2 times 4C2 +
6C1 times 5C3 +
6C4
= 240 ways
Atleast 1 card repeating = total - No cards repeating.
Atleast 1 card repeating = 495 - 240
P(the above event) = (495 - 240) / 495. Choice C
Look forward to a better method or pls highlight if there are some issues here.
Thanks.
Interesting problem ... Perhaps amongst the tough ones in GMAT.
Atleast 1 card repeating = total - No cards repeating.
Total card possibilities = 12 C 4 = 495
No card repeatings = 5 ways of achieving this
4 cards from deck1 (or)
3 cards from d1 and 1 of the other cards from d2 (or)
2 cards from d1 and 2 of the other cards from d2 (or)
1 card from d1 and 3 of the other cards from d2 (or)
4 cards from d2
=
6C4 +
6C3 times 3C1 +
6C2 times 4C2 +
6C1 times 5C3 +
6C4
= 240 ways
Atleast 1 card repeating = total - No cards repeating.
Atleast 1 card repeating = 495 - 240
P(the above event) = (495 - 240) / 495. Choice C
Look forward to a better method or pls highlight if there are some issues here.
Thanks.
Last edited by 4GMAT_Mumbai on Thu Jul 08, 2010 11:34 pm, edited 1 time in total.
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Let p be the probability that Bill finds no pair of cards that have the same value. Then probability that Bill finds at least one pair of cards that have the same value = 1- p
Total no. of ways in which 4 cards can be drawn from 12 cards = 12C4 = 495
Number of ways in which Bill finds no pair of cards with the same value:
There are 6 cards within a suit and we want to select 4 cards. So, this can be done in 6C4 = 15 ways.
For 15 values each card can be of either suit.
1st card can be chosen from 2 suits, similarly 2nd, 3rd & 4th card can be chosen from 2 suits. We can do this 2^4 = 16 ways
So, number of ways to choose 4 cards so that no pairs exist = 15×16 = 240
So, p = 240/495 = 16/ 33
Hence 1- p = 1-16/33 =17/33
[spoiler]The correct answer is (C).[/spoiler]
Total no. of ways in which 4 cards can be drawn from 12 cards = 12C4 = 495
Number of ways in which Bill finds no pair of cards with the same value:
There are 6 cards within a suit and we want to select 4 cards. So, this can be done in 6C4 = 15 ways.
For 15 values each card can be of either suit.
1st card can be chosen from 2 suits, similarly 2nd, 3rd & 4th card can be chosen from 2 suits. We can do this 2^4 = 16 ways
So, number of ways to choose 4 cards so that no pairs exist = 15×16 = 240
So, p = 240/495 = 16/ 33
Hence 1- p = 1-16/33 =17/33
[spoiler]The correct answer is (C).[/spoiler]
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On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
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Since the problem asks us to find atleast one pair of cards that have the same value, a simpler approach would be to find the probability if none of the cards have the same value and subtract that number from 1.paridhi wrote:Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
(A) 8/33
(B) 62/165
(C) 17/33
(D) 103/165
(E) 25/33
Can someone please tell me the approach to do this question?
P(A) = 1 - P(~A)
Thus, P(finding one pair) = 1 - P(not finding a single pair)
P(~Single Pair):
Calculate the denominator as follows:
4 cards can be chosen out of 12 cards in 12C4 = (12*11*10*9)/4! ways
Calculate the numerator as follows:
Since nCr = (nPr)/r!, calculate an ordered pair(permutation) and divide it by r!
1st card can be chosen in 12 ways
2nd------------------------ in 10 ways(if 2 is chosen, the other 2 in the deck cannot be chosen)
3rd------------------------- in 8 ways
4th--------------------------in 6 ways
Thus, numerator = (12*10*8*6)/4!
Thus, P(~Single pair) = num/den = (12*10*8*6)/(12*11*10*9) = 16/33
Thus P(Atleast one pair) = 1-16/33 = 17/33
Hope it helps,
meshtrap