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Probability - I got this wrong in one of practice tests !!

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Probability - I got this wrong in one of practice tests !!

by selfmade » Fri Aug 13, 2010 9:59 am
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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Aiming for 780
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by kvcpk » Fri Aug 13, 2010 10:19 am
selfmade 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
My method might be a bit lengthy. But I would still post it :)

We need to find the chance for atleast one pair.

I am finding the chance for no pair.
We need to pick 4 cards from 2 sets of 6 each.
This can be doen in:
All 4 from first set = 6c4 = 15 ways
3 from 1st set and 1 from second set = 6c3 * 3c1 =60 ways (3C1 Because, the pairing cards should not be chosen)
2 from 1st set and 2 from second set. = 6c2 * 4c2 = 90 ways
1 from first set and 3 from second set = 6C1 * 5c3 = 60 ways
All 4 from second set = 6c4 = 15 ways
Hence total = 240.

Now, total number of ways of choosing 4 cards from the 12 cards is 12c4 = 495

Hence chance of getting atleast 1 pair = 1-(240/495) = 17/33

Hope this helps!!
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by uwhusky » Fri Aug 13, 2010 10:27 am
Can someone point out the flaw to my approach, which is wrong because I cannot find the answer for it.

12 cards, only 2 can have the same value. My approach is to find the odd which all 4 cards won't make a pair.

12/12 * 10/11 * 9/10 * 8/9 = 8/11

1 - 8/11 = 3/11.

What's wrong with this approach?
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by CompBanker » Fri Aug 13, 2010 12:16 pm
uwhusky wrote:Can someone point out the flaw to my approach, which is wrong because I cannot find the answer for it.

12 cards, only 2 can have the same value. My approach is to find the odd which all 4 cards won't make a pair.

12/12 * 10/11 * 9/10 * 8/9 = 8/11

1 - 8/11 = 3/11.

What's wrong with this approach?
Your solution is incorrect because you're not taking into account that the 2nd + 3rd + 4th cards could produce a match with each other. All in all, you're very much on the right track. My solution would be:

1 - (12/12)(10/11)(8/10)(6/9) = 17/33.

In this solution, you're trying to figure out what the odds of NO cards matching and then subtracting that probability from 1. So basically:

100% (Odds of pulling the 1st card)
10/11 (Odds of pulling the single card of the remaining 11 that matches the first one).
8/10 (Odds of pulling either of the two cards of the remaining 10 that match either the 1st or the 2nd).
6/9 (Odds of pulling one of the three cards of the remaining 9 that match either the 1st, 2nd, or the 3rd).

CompBanker
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by uwhusky » Fri Aug 13, 2010 5:24 pm
Thank you for the information.

I didn't think I'll be able to utilize kvcpk's approach on test day.
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by kvcpk » Sat Aug 14, 2010 12:04 am
uwhusky wrote:Thank you for the information.

I didn't think I'll be able to utilize kvcpk's approach on test day.
I thought the same too :)
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don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
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by sanju09 » Sat Aug 14, 2010 12:53 am
selfmade 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

This seems favorite to many, discussed so many times, already

https://www.beatthegmat.com/deck-of-cards-t28564.html

https://www.beatthegmat.com/combination- ... t8924.html

https://www.beatthegmat.com/probability- ... 61115.html

https://www.beatthegmat.com/probability- ... 32580.html
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