Bill has a set of 6 black cards and a set of 6 red cards

ANSWER: B
Ways to select one pair out of 6 pairs: 6C1=6
Ways to select 2 cards out of remaining 10 cards such that no pair is drawn: 10C2-5. Please note 5 has to be subtracted for possible 5 pairs that were included in 10C2.
Required probability: (6*(10C2-5))/12C4 =(6*(45-5))/12C4=16/33

Alto:
One pair can be selected out of 6 pairs in 6C1 ways.
Probability of 1 pair thus selected and 2 different number cards: (12/12*1/11*10/10*8/9)=8/9*11
Probability of all six pairs: 6*8/9*11=16/33

Alto:
First, recognize that P(at least one pair) = 1 - P(no pairs)
P(no pairs) = P(select any 1st card AND select any non-matching card 2nd AND select any non-matching card 3rd AND select any non-matching card 4th)
= P(select any 1st card) x P(select any non-matching card 2nd) x P(select any non-matching card 3rd) x P(select any non-matching card 4th)
= 1 x 10/11 x 8/10 x 6/9
= 16/33

So, P(at least one pair) = 1 - 16/33
= 17/33

Now, P(at least one pair)=P(exactly one pair)+P(two pairs)
P(two pairs)= 6C2/12C4=1/33
Now, P(exactly one pair)=17/33-1/33=16/33

We can solve this question using probability rules.

First, recognize that P(at least one pair) = 1 - P(no pairs)

P(no pairs) = P(select any 1st card AND select any non-matching card 2nd AND select any non-matching card 3rd AND select any non-matching card 4th)
= P(select any 1st card) x P(select any non-matching card 2nd) x P(select any non-matching card 3rd) x P(select any non-matching card 4th)
= 1 x 10/11 x 8/10 x 6/9
= 16/33

So, P(at least one pair) = 1 - 16/33
= 17/33
= C

Cheers,
Brent

This is a copy of the following MGMAT question: https://gmatclub.com/forum/bill-has-a-s ... 57417.html
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Everyone is forgetting to account for the possibility of two pairs. Everyone says "P(exactly one pair) = 1 - P(No pairs). Does anyone know how to do it with P(Exactly one pair) = 1 - (P(no pairs)*P(2 pairs)?

This is my math.

P(No Pairs) = (12/12) * (10/11) * (8/10) * (6/9) = 16/33
P(Two Pairs) = (12/12) * (1/11) * (10/10) * (1/9) = 1/99

Then I multiplied them both and got roughly 1/198.

I feel like everyone forgets the possibility of there being two pairs. Taking that into account, is this correct?

**EDIT**
I retried this problem. Here is my new math.

"Probability of exactly 1 pair" is similar to getting A - A - B - C in which A,B & C are all digits 1-6.
So the Probability of getting A or any number for that matter is (1)
Probability of matching that same number A is (1/11)
Probability of getting any other number B is (1)
Probability of getting any other number that doesn't match B is C = (8/9)
Hence, (1)*(1/11)*(1)*(8/9)
Then you need to account for all the possible permutations of A - A - B - C which is (4!/2!)
Therefore, Probability of getting exactly 1 pair =
(1)*(1/11)*(1)*(8/9)*(4)*(3) = 8/11

Please let me know if this is correct!!

Alternate Solution:

Ways to select one pair out of 6 possibilities (1-1,2-2,3-3,4-4,5-5,6-6): 6C1
Now the remaining 2 cards cannot be a pair, so the 1st of them will have 5C1 possibility (since one number is already taken up in the pair cards).
2nd non-pair card will have 4C1 possibility (as 1 is gone with pair card, and it cannot be same as the 1st non-pair card)

But, you also need to consider for colors, so non-pair numbers can have reverse so we multiply by 2.
Hence total ways: 6C1 x 5C1 x 4C1 x 2
And total possibilities: 12C4

Answer: (6C1 x 5C1 x 4C1 x 2) / 12C4 = 16/33

please some one provide clearly how to fin out exactly one pair. don't confuse us