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Probability - pairs of cards

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sachin2411: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Fri Dec 03, 2010 11:25 am

by sachin2411 » Tue Jul 26, 2011 2:24 am

A question from MGMAT CAT:

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

OA: (C) 17/33

This is how i calculated:
No. of ways of selecting a pair = 6C1

Once i have a pair, i am left with 10 cards,
So number of ways of selecting 2 from 10 = 10C2

So Probability = (6C1 X 10C2) / 14C4

Please let me know, what is wrong with my approach.

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Source: Beat The GMAT — Problem Solving | Tue Jul 26, 2011 2:24 am

top_business_2011: Senior | Next Rank: 100 Posts; Posts: 47; Joined: Wed Sep 03, 2008 11:18 pmFollowed by: 1 member; Thanked: 5 times

by top_business_2011 » Tue Jul 26, 2011 3:39 am

sachin2411 wrote:A question from MGMAT CAT:

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

OA: (C) 17/33

This is how i calculated:
No. of ways of selecting a pair = 6C1

Once i have a pair, i am left with 10 cards,
So number of ways of selecting 2 from 10 = 10C2

So Probability = (6C1 X 10C2) / 14C4 [You considered only one pair, while the question is about at least one pair][ Note also that the 10C2 may or may not hold pairs, rendering your solution incorrect]

Please let me know, what is wrong with my approach.

When dealing with probability problems that are framed as 'at least' or 'at most', sometimes it's easier to find probability of a complement.That is, if it says 'at least 1', then find the probability of zero and subtract the value from 1.[ Because, the total probability is 1.]

In our problem, that's exactly what we have. Hence, let's find probability of zero number of pairs.

Have four slots in which you shall put maximum possibilities.
(12 * 10 * 8 * 6) This is the total number of combinations in which we have no pair with similar values. Let me explain this: The first 12 tells us that we can choose any card as our first card; whereas, the second is 10 because we can't pick the already selected card and its similar value in the other suit. The third is 8 because in addition to the previous two, we have other two cards out of consideration. The same holds true for the fourth 6.

Now, divide that by (12 *11*10*9), which is the total number of combinations.You shall have 16/33.

Because we calculated the value for the complement, we need to subtract 16/33 from 1. And that gives us 17/33.

The answer is C.

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sachin2411: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Fri Dec 03, 2010 11:25 am

by sachin2411 » Tue Jul 26, 2011 3:45 am

YAMLAKSIRA wrote:
sachin2411 wrote:A question from MGMAT CAT:

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

OA: (C) 17/33

This is how i calculated:
No. of ways of selecting a pair = 6C1

Once i have a pair, i am left with 10 cards,
So number of ways of selecting 2 from 10 = 10C2

So Probability = (6C1 X 10C2) / 14C4 [You considered only one pair, while the question is about at least one pair][ Note also that the 10C2 may or may not hold pairs, rendering your solution incorrect]

Please let me know, what is wrong with my approach.
Well, 10C2 can be anything, either its a pair of 2 equal card or two different cards. At least 1 pair thing is already dealt by 6C1, for rest 2 it can be anything so 10C2.

Please help.

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Frankenstein: Legendary Member; Posts: 1448; Joined: Tue May 17, 2011 9:55 amFollowed by: 53 members; Location: India; Thanked: 375 times

by Frankenstein » Tue Jul 26, 2011 4:26 am

sachin2411 wrote:A question from MGMAT CAT:

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

OA: (C) 17/33

This is how i calculated:
No. of ways of selecting a pair = 6C1

Once i have a pair, i am left with 10 cards,
So number of ways of selecting 2 from 10 = 10C2

So Probability = (6C1 X 10C2) / 14C4

Please let me know, what is wrong with my approach.

Hi,
First, your denominator is a problem. I guess you meant 12C4.
Coming to the numerator
Let the 2 suits be A1,A2,...,A6 and B1,B2,...B6.
Now, you are selecting a pair in 6C1 ways, fine
While selecting the other two from the other 10, you are using 10C2.
So, in essence you are actually counting certain pairs twice.
Let's say your first pair is A1,B1
The other pair can be all possible pairs from A2,....,A6,B2,...B6.
So, when your first pair is A1,B1, you are counting (A2,B2),(A3,B3)...(A6,B6) as the second pair.
Similarly, When you first pair is (A2,B2), you are again counting (A1,B1),(A3,B3),...(A6,B6) as second pair. So, (A1,B1)and (A2,B2) is counted twice.
Similarly, When you first pair is (A3,B3), you are again counting (A1,B1),(A2,B2),...(A6,B6) as second pair. So, [(A1,B1),(A3,B3)] and [(A2,B2),(A3,B3)] are counted twice.
Similarly in all 6 cases, you are counting 6C2 pairs twice.
So, your numerator should be 6C1*10C2 - 6C2 = 6*9*5 - 15 = 255
So, probability is 255/12C4 = 255*24/12.11.10.9 = 17/33

Hence, C

If you need other approaches for this question, you can follow:
https://www.beatthegmat.com/probability- ... 85056.html

Cheers!

Things are not what they appear to be... nor are they otherwise

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sachin2411: Junior | Next Rank: 30 Posts; Posts: 17; Joined: Fri Dec 03, 2010 11:25 am

by sachin2411 » Tue Jul 26, 2011 6:30 am

Hey, thanks a lot mate..

now i figured it out...!!!

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