Bill has a small deck of 12 playing cards made up of only 2 suits of 6

We can use the equation:

1 = P(at least one pair of cards in 4) + P(no pairs of cards in 4)

Let’s determine the probability of not finding a pair of cards when 4 are flipped.

P(first flip) = 1

P(second flip is not a pair with the first card) = 10/11

P(third card is not a pair with the first or second card) = 8/10

P(fourth card is not a pair with the first, second, or third card) = 6/9

Thus, the probability of not selecting a pair is:

10/11 x 8/10 x 6/9

10/11 x 4/5 x 2/3

2/11 x 4 x 2/3

16/33

Since 1 = P(at least one pair of cards in 4) + P(no pairs of cards in 4),

P(at least one pair of cards in 4) = 1 - P(no pairs of cards in 4) = 1 - 16/33 = 17/33.

Answer: C

Heres a simple and cool and innovative method.

Bill chooses 2 cards with at least 1 pair. Therefore the problem can be solved in the following manner:-

a) total possibilities = 12x11x10x9 cards
b) If he choose at least 1 pair, then there are 2
Case1: Where he chooses exactly 1 pair = 12x10x8x3 (Using the Bunuel method, I simply replaced 6 by 3 because at least 1 of the 3 numbers is
Case2: Where he chooses exactly 2 pairs = 12x10x2x1 (Uptil 10, then he has only 2 cards)

Probability = (12x10x8x3 + 12x10x2x1)/12x11x10x9 = 26/99.

But the answer is not correct. I fail to understand why? I have 25 short on the numerator.

Can someone please explain the 2^4 part in original solution.

If Say, I have to choose 2 numbers from 5, say the chosen numbers are 1 and 2 , then the way I can include is (1,2) (2,1) I can not include (1,1) and (2,2) how can there be four ways to get 1 and 2 from 2 suits(someone gave this ans as a solution of selecting one pair by 6c1*5c2x2^2) where I am assuming that method of getting 2^2 is same as getting 2^4.

Help will be appreciated... Thanks

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Posted from my mobile device

You are looking for the ways in which you have no pair. So you will select 4 numbers out of 6 (from 1 to 6). This is done in 6C4 ways. Say we get 1, 2, 4, 5 in one case.
But we have 2 suits say spades and hearts. The 1 can be of either spades or hearts (it will lead to 2 different cases). Similarly, 2 can be either spades or hearts and so on...
So for each number, you have 2 options of the suit. Hence you multiply by 2 * 2 * 2 * 2 = 2^4.

Hi VeritasPrepKarishma
Thanks for the reply. Like in your example you mentioned 1,2,4,5.so as u rightly mentioned that each could come from different suits. So aren't we counting (1,1) here which is not required.

We have already eliminates the probability of getting (3,3) (6,6) likewise should we subtract some number from 2^4.

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Posted from my mobile device

No, we cannot have (1, 1). Note that we have 4 distinct numbers already - 1, 2, 4 and 5. Each will appear once only.
But we can have two cases -
1 of spades, 2 of spades, 4 of hearts and 5 of spades
1 of hearts, 2 of spades, 4 of hearts and 5 of spades

So by multiplying by 2, we are accounting for these 2 cases. Same goes for each number. That is why we have 2*2*2*2

I think there is a different way to solve this Question.
12C4 = 495

at least 1 pair = 2c2 *6 * 10c2 -5 = 6*(45-5) = 240 ( -5 for to avoid any remaining pair)
At least 2 pair: 6c2 = 15

Total possibilities = 15 + 240=255

255/495 = 17/33

Dear IanStewart EMPOWERgmatRichC VeritasKarishma Bunuel,

Thank you for your previous replies

To have a clear understanding, I would like to know how to calculate this question DIRECTLY using Probability method.

Probability Method: Prob of 1 pair + Prob of 2 pairs

Is this approach correct?

(12/12)(1/11)(10/10)(8/9) * (4!/2!2!) + (12/12)(1/11)(10/10)(1/9) * (4!/2!2!2!) = 17/33

I understand that order DOES matter when calculating using Probability Method.
Hence I multiply the first part with 4!/2!2! because of arranging 4 objects with 1 set of 2 repeats; however, the order of 2 distinct objects does not matter. Or another way to look is that I pick 2 positions for the pair out of 4 positions.

and

(this one is very tricky) the second part with 4!/2!2!2! because of arranging 4 objects with 2 sets of 2 repeats each; however, the order of 2 sets itself does not matter.

I'm very unsure of the highlighted portion (In fact, I find the combinatorics involved very complicated. Do you have any advanced principles / formula for these kinds of combinations?)
Can you please share how to correctly use direct Probability Method in this question?

(I know that this method is tricky, but I would like to have a clear understanding of Probability Method)

In general, what is your approach in solving probability? - Probability Method or Combinatoric Method?
Please help!
Thank you

You are unnecessarily complicating it far too much.

One pair: (12/12)(1/11)(10/10)(8/9) * 4C2 = 16/33
(4C2 selects the two spots in which the pair appears)

Two pairs: (12/12)(1/11)(10/10)(1/9) * (4C2 / 2) = 1/33
4C2 selects the two spots in which one pair appears but the other two spots are also a pair. The two pairs are not distinct so we divide further by 2.

Total = 17/33

I couldnt understand the part “10c2- the number of repetitions”
Can we not use
6C1 x 10C1 x 8C1? Considering that we select 1 pair out of 6 and then 1 card from the remaining 10 and 1 card from remaining 8 pair?

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Let's find the probability of "losing" and then take 1-losing to find "winning."

What's the probability that he gets SOMETHING on the first card? 1/1.
There are now 11 cards left. How many of them have a number that's different from the first number? 10. So, what's the probability he gets a different number with the second card? 10/11.
There are now 10 cards left. How many of them have a number that's different from both the first number and the second number? 8. So, what's the probability he gets a different number with the third card? 8/10.
There are now 9 cards left. How many of them have a number that's different from the first number, second number, and third number? 6. So, what's the probability he gets a different number with the fourth card? 6/9.

(1/1)*(10/11)*(8/10)*(6/9) = 16/33

1-(16/33) = 17/33

Answer choice C.

Given: 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.

Asked: What is the chance that Bill finds at least one pair of cards that have the same value?

Total ways to select 4 cards = 12C4 = 11*5*9 = 495
Number of ways to select 4 cards in at least 1 pair = Total ways to select 4 cards - Number of ways to select 4 cards with no pair = 12C4 - 12*10*8*6/4! = 255

Probability that finds at least one pair of cards that have the same value = 255/495 = 17/33

IMO C

I got the right answer by doing the complementary approach, like Bunuel "another way"-approach:
prob(XYZW)=(12/12)*(10/11)*(8/10)*(6/9)=16/33
1-prob(XYZW)=17/33

But how can i calculate the number directly, instead of doing the complementary probability?

I tried do this, but im missing something... can someone see what is wrong?

there are 4 possible configurations (where X != Y != Z):
i) prob(YXXZ)=(12/12)*(10/12)*(1/10)*(8/9) = 8/99
ii) prob(YZXX)=8/99
iii) prob(XXYZ)=8/99
iv) prob(XXYY)=(12/12)*(1/11)*(10/10)*(1/9)=1/99

prob(at least 1 pair) = (3*prob(YZXX)+prob(XXYY))=3*(8/99)+(1/99)=25/99