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Probability | Playing Cards

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Probability | Playing Cards

by jsockol » Mon May 21, 2012 9:07 am
I am looking to solve the below using the fundamental counting principle. I understand how to calculate the problem intuitively -- more concerned with the mathematical approach for counting the numerator in particular

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
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by Stuart@KaplanGMAT » Mon May 21, 2012 11:42 am
Hi! We want the probability of getting at least one matching pair. So, the only thing we don't want to happen is 0 matching pairs. Accordingly, this is a great question on which to apply the "one minus" approach.

Remember this key formula for complex probability problems:

Prob(want) = 1 - Prob(don't want)

So, let's calculate the probability of getting no matches:

1st card: 12/12 cards are fine
2nd card: 10/11 cards are fine (only 1 of the remaining 11 will match what we've already chosen)
3rd card: 8/10 cards are fine (2 of the remaining 10 will match one of our first two picks)
4th card: 6/9 cards are fine (3 of the remaining 9 will match one of our first three picks)

So, the prob(no matches) = 12/12*10/11*8/10*6/9

Ignoring the 12/12 (which is just 1):

10/11 * 8/10 * 6/9

= 6*8*10/9*10*11

cancelling out the 10s:

= 48/99

dividing by 3:

= 16/33

Now, this question designer was kind and gentle; on the actual GMAT, 16/33 would almost always be one of the choices. Remember, we're not done - we need to use the "one minus" equation:

Prob(want) = 1 - 16/33 = 17/33... choose C!
jsockol wrote:I am looking to solve the below using the fundamental counting principle. I understand how to calculate the problem intuitively -- more concerned with the mathematical approach for counting the numerator in particular

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
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by Anurag@Gurome » Mon May 21, 2012 8:24 pm
jsockol wrote:I am looking to solve the below using the fundamental counting principle. I understand how to calculate the problem intuitively -- more concerned with the mathematical approach for counting the numerator in particular

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
Number of ways to choose 4 different cards out of 6 different values = 6C4
Now each of the 4 chosen cards can be of 2 different suits, so this can be done is 2^4 ways
Total number of ways to choose 4 cards out of 12 = 12C4

Probability that Bill does not find even one pair of cards that have the same value = (6C4 * 2^4)/12C4 = 16/33

Therefore, required probability = 1 - 16/33 = [spoiler]17/33[/spoiler]

The correct answer is C.
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by jsockol » Tue May 22, 2012 3:22 pm
Anurag@Gurome wrote:
jsockol wrote:I am looking to solve the below using the fundamental counting principle. I understand how to calculate the problem intuitively -- more concerned with the mathematical approach for counting the numerator in particular

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
Number of ways to choose 4 different cards out of 6 different values = 6C4
Now each of the 4 chosen cards can be of 2 different suits, so this can be done is 2^4 ways
Total number of ways to choose 4 cards out of 12 = 12C4

Probability that Bill does not find even one pair of cards that have the same value = (6C4 * 2^4)/12C4 = 16/33

Therefore, required probability = 1 - 16/33 = [spoiler]17/33[/spoiler]

The correct answer is C.
Thanks again -- extremely helpful. Had been struggling on coming up with the 2^4 concept on my own. Much appreciated
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