Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, 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?
8/33
62/165
17/33
103/165
25/33
OA C
My Approach:
Probability of finding one pair + probability of finding two pairs
=> (12 x 1 x 10 x 8 + 12 x 1 x 10 x 1)/12C4
Can somehelp please help me understand what wrong did I do?
Also, I saw OE and see that the approach is to calculate probability of individual outcome, i.e, 1 st card drawn has probabibility 1 and the second 10/11 and so on. Although, I understand this approach, I am not sure when to apply this for easy calculations. Can some please help?
MGMAT CAT Probability and Combinatronics
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 bubbliiiiiiii
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.
Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, 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
Total number of cases to draw 4 cards from 12 cards > 12C4
cases of a number chosen as pair : 6
among remained cards: different colors and different numbers > 5C1*4C1
same colors > 5C2*2
So the number of cases to draw only one pair of cards > 6*(5C1*4C1 + 5C2 *2)
The number of cases to draw two pairs of cards > 6C2
The probability > (6C2 + 6*(5C1*4C1 + 5C2 *2))/12C4 = (3*5 + 3*5*16)/(11*5*9)
= 17/33
The answer, therefore, C
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Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, 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
Total number of cases to draw 4 cards from 12 cards > 12C4
cases of a number chosen as pair : 6
among remained cards: different colors and different numbers > 5C1*4C1
same colors > 5C2*2
So the number of cases to draw only one pair of cards > 6*(5C1*4C1 + 5C2 *2)
The number of cases to draw two pairs of cards > 6C2
The probability > (6C2 + 6*(5C1*4C1 + 5C2 *2))/12C4 = (3*5 + 3*5*16)/(11*5*9)
= 17/33
The answer, therefore, C
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Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.
Hi bubbliiiiiiii
I've thought your approach.
Your approach showed that 12 x 1 x 10 x 8 is the probability of finding one pair. But it included some of repetitions of cases; by example your approach included (B1, W1, B2, W3), (W1, B1, B2, W3), (B1, W1, W3, B2), (W1, B1, W3, B2).
Even if all above four cases represents same outcome, you counted them all. So you should divide 12 x 1 x 10 x 8 with 4.
Similarly counting the cases of two pairs, you have counted the repetitions.
(B1, W1, B2, W2), (W1, B1, B2, W2), (B1, W1, W2, B2), (W1, B1, W2, B2),
(B2, W2, B1, W1), (B2, W2, W1, B1), (W2, B2, B1, W1), (W2, B2, W1, B1)
Similarly should divide 12 x 1 x 10 x 1 with 8.
I hope it would be helpful to you. Good luck to you
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Hi bubbliiiiiiii
I've thought your approach.
Your approach showed that 12 x 1 x 10 x 8 is the probability of finding one pair. But it included some of repetitions of cases; by example your approach included (B1, W1, B2, W3), (W1, B1, B2, W3), (B1, W1, W3, B2), (W1, B1, W3, B2).
Even if all above four cases represents same outcome, you counted them all. So you should divide 12 x 1 x 10 x 8 with 4.
Similarly counting the cases of two pairs, you have counted the repetitions.
(B1, W1, B2, W2), (W1, B1, B2, W2), (B1, W1, W2, B2), (W1, B1, W2, B2),
(B2, W2, B1, W1), (B2, W2, W1, B1), (W2, B2, B1, W1), (W2, B2, W1, B1)
Similarly should divide 12 x 1 x 10 x 1 with 8.
I hope it would be helpful to you. Good luck to you
www.mathrevolution.com
 The oneandonly World's First Variable Approach for DS and IVY Approach for PS that allow anyone to easily solve GMAT math questions.
 The easytouse solutions. Math skills are totally irrelevant. Forget conventional ways of solving math questions.
 The most effective time management for GMAT math to date allowing you to solve 37 questions with 10 minutes to spare
 Hitting a score of 45 is very easy and points and 4951 is also doable.
 Unlimited Access to over 120 free video lessons at https://www.mathrevolution.com/gmat/lesson
Our advertising video at https://www.youtube.com/watch?v=R_Fki3_2vO8
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We can solve this question using probability rules.Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, 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
First, recognize that P(at least one pair) = 1  P(no pairs)
P(no pairs) = P(select any 1st card AND select any nonmatching card 2nd AND select any nonmatching card 3rd AND select any nonmatching card 4th)
= P(select any 1st card) x P(select any nonmatching card 2nd) x P(select any nonmatching card 3rd) x P(select any nonmatching 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
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We can also solve the question using counting techniques.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, there are 2 cards in the deck that have the same 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
First recognize that P(at least one pair) = 1  P(no pairs)
We'll find P(no pairs)
First, the number of possible outcomes.
We have 12 cards and we select 4.
This is accomplished in 12C4 ways = 495
Aside: If anyone is interested, we have a free video on calculating combinations (like 12C4) in your head: https://www.gmatprepnow.com/module/gmatcounting?id=789
Now we count the number of ways to select 4 different cards with no pairs. In other words, we want 4 different card values.
First look at how many different cards we can select (WITHOUT considering the suits)
There are 6 card values and we want to select 4.
This is accomplished in 6C4 = 15 ways.
For each of these 15 values (e.g., 1,2,4,5) each card can be of either suit.
So, for example, in how many ways can we have card values of 1,2,4,5?
For the 1card we have 2 suits from which to choose.
For the 2card we have 2 suits from which to choose.
For the 4card we have 2 suits from which to choose.
For the 5card we have 2 suits from which to choose.
So, for the selection 1,2,4,5 there are 2x2x2x2=16 possibilities.
So, the number of ways to select 4 cards such that there are no pairs is 15x16=240
So, the probability that there are no pairs = 240/495 = 16/33
So, P(at least one pair) = 1 16/33 = 17/33 = C
Cheers,
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I think I've got a succinct solution for you!
You had the right idea, you just need to make sure that you're finding (# with exactly one pair) + (# with exactly two pairs).
(# with exactly one pair) = (one pair) * (one nonpair)
= (any of the 6 pairs) * (any 2 of the other 10 cards EXCEPT one of the other 5 pairs)
= 6 * ((10 choose 2)  5)
= 6 * (45  5)
= 240
(# with exactly two pairs) = (any two of the six possible pairs)
= (6 choose 2)
= 15
So there are 255 possibilities. 255 / (12 choose 4) = 17/33, and we're done!
You had the right idea, you just need to make sure that you're finding (# with exactly one pair) + (# with exactly two pairs).
(# with exactly one pair) = (one pair) * (one nonpair)
= (any of the 6 pairs) * (any 2 of the other 10 cards EXCEPT one of the other 5 pairs)
= 6 * ((10 choose 2)  5)
= 6 * (45  5)
= 240
(# with exactly two pairs) = (any two of the six possible pairs)
= (6 choose 2)
= 15
So there are 255 possibilities. 255 / (12 choose 4) = 17/33, and we're done!
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Hi All,
Thanks for responses having alternate approaches.
My approach was this:
Probability = #Favourable outcomes/#Total outcomes
=> Combinations having atleast one pair/Total no. of combinations.
Combinations having atleast one pair I calculated as under:
There 12 cards in all (1B, 2B, 3B, 4B, 5B, 6B, 1R, 2R, 3R, 4R, 5R, 6R)
#Number of outcomes having one pair only
Number of ways I can choose the first card out of 12 cards = 12C1 = 12
Number of ways I can choose the same card from the remaining set of 11 cards = 1
Number of ways I can choose third card out of remaining 10 cards = 10C1 = 10
Number of ways I can choose fourth card out of remaining 8 cards = 8C1 = 8
#Number of outcomes having two pairs
Number of ways I can choose 1 card out of 12 cards = 12C1 = 12
Number of ways I can choose the 2 card to form a pair = 1
Number of ways I can choose 3 card out of remaining 10 cards = 10C1 = 10
Number of ways I can choose 4 card to form a pair with third card = 1
Thus, total favourable outcomes = One pair + Two pairs
=> 12 X 1 X 10 X 8 + 12 X 1 X 10 X 1
Total number of outcomes = 12C4
Probability = (12 X 1 X 10 X 8 + 12 X 1 X 10 X 1)/12C4 .. which is incorrect.
I think I have either repeated a combination several times. To correct, dividing each set by 4! to remove order prerence but still no benefit!
Can someone please help me correct the above approach?
Matt/Brent/Max  Thanks again for helping with alternate solutions. Can you also help me evaluate my approach?
Thanks for responses having alternate approaches.
My approach was this:
Probability = #Favourable outcomes/#Total outcomes
=> Combinations having atleast one pair/Total no. of combinations.
Combinations having atleast one pair I calculated as under:
There 12 cards in all (1B, 2B, 3B, 4B, 5B, 6B, 1R, 2R, 3R, 4R, 5R, 6R)
#Number of outcomes having one pair only
Number of ways I can choose the first card out of 12 cards = 12C1 = 12
Number of ways I can choose the same card from the remaining set of 11 cards = 1
Number of ways I can choose third card out of remaining 10 cards = 10C1 = 10
Number of ways I can choose fourth card out of remaining 8 cards = 8C1 = 8
#Number of outcomes having two pairs
Number of ways I can choose 1 card out of 12 cards = 12C1 = 12
Number of ways I can choose the 2 card to form a pair = 1
Number of ways I can choose 3 card out of remaining 10 cards = 10C1 = 10
Number of ways I can choose 4 card to form a pair with third card = 1
Thus, total favourable outcomes = One pair + Two pairs
=> 12 X 1 X 10 X 8 + 12 X 1 X 10 X 1
Total number of outcomes = 12C4
Probability = (12 X 1 X 10 X 8 + 12 X 1 X 10 X 1)/12C4 .. which is incorrect.
I think I have either repeated a combination several times. To correct, dividing each set by 4! to remove order prerence but still no benefit!
Can someone please help me correct the above approach?
Matt/Brent/Max  Thanks again for helping with alternate solutions. Can you also help me evaluate my approach?
Regards,
Pranay
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bubbliiiiiiii wrote:#Number of outcomes having one pair only
Number of ways I can choose the first card out of 12 cards = 12C1 = 12
Number of ways I can choose the same card from the remaining set of 11 cards = 1
Number of ways I can choose third card out of remaining 10 cards = 10C1 = 10
Number of ways I can choose fourth card out of remaining 8 cards = 8C1 = 8
This is right, but you need to divide both of these pairs by 2. The cards that match should be (12 * 1)/2, and the cards that don't should be (10*8)/2. So you should have (12 * 1 * 10 * 8) / (2 * 2), or 240.
#Number of outcomes having two pairs
Number of ways I can choose 1 card out of 12 cards = 12C1 = 12
Number of ways I can choose the 2 card to form a pair = 1
Number of ways I can choose 3 card out of remaining 10 cards = 10C1 = 10
Number of ways I can choose 4 card to form a pair with third card = 1
Same idea here, at least to start. (12 * 1 * 10 * 1) / (2 * 2) = 30. Then you've got to divide by 2 again, because you're considering the ORDER of the pairs. (For instance, you're treating 6 and 5 as distinct from 5 and 6, but they aren't distinct for our purposes.) So we divide by 2 again, yielding 15.

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As you've certainly gathered by now, the hell of combinatorics is trying to keep track of what you may have overcounted! One way to avoid this, I think, is to be as neat as you can about the groups, as I did in my first solution. This keeps you from having to do too much division, and makes things slightly easier.
But in my experience, combinatorics is the only basic branch of math in which I feel like I get worse the more I do. My instincts for overcounting haven't improved, despite doing all sorts of neato, beyond the GMAT combinatorial problems. I still stink and make bazillions of elementary mistakes. > <
But in my experience, combinatorics is the only basic branch of math in which I feel like I get worse the more I do. My instincts for overcounting haven't improved, despite doing all sorts of neato, beyond the GMAT combinatorial problems. I still stink and make bazillions of elementary mistakes. > <
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Hi Matt,
Thanks for the correction.
Mistake I did was to divide it by 4! rather than 4.
Regards,
Pranay
Thanks for the correction.
Mistake I did was to divide it by 4! rather than 4.
Regards,
Pranay
Regards,
Pranay
Pranay
 bubbliiiiiiii
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[email protected] wrote:As you've certainly gathered by now, the hell of combinatorics is trying to keep track of what you may have overcounted! One way to avoid this, I think, is to be as neat as you can about the groups, as I did in my first solution. This keeps you from having to do too much division, and makes things slightly easier.
But in my experience, combinatorics is the only basic branch of math in which I feel like I get worse the more I do. My instincts for overcounting haven't improved, despite doing all sorts of neato, beyond the GMAT combinatorial problems. I still stink and make bazillions of elementary mistakes. > <
Regards,
Pranay
Pranay
P(no pairs) = P(select any 1st card AND select any nonmatching card 2nd AND select any nonmatching card 3rd AND select any nonmatching card 4th)
= P(select any 1st card) x P(select any nonmatching card 2nd) x P(select any nonmatching card 3rd) x P(select any nonmatching card 4th)
= 1 x 10/11 x 8/10 x 6/9
Hi Brent ,
Can you please explain the above part?
Specially this 1 x 10/11 x 8/10 x 6/9 how did we get this no.?
Many thanks in advance.
SJ
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P(at least 1 pair) = 1  P(no pairs).Bill has a set of 6 black cards and a set of 6 red cards. Each card has a number from 1 through 6, such that each of the numbers 1 through 6 appears on 1 black card and 1 red card. Bill likes to play a game in which he shuffles all 12 cards, 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
P(no pairs):
Any of the 12 cards can be selected first.
P(2nd card does not match the first) = 10/11. (Of the 11 remaining cards, any but the mate of the 1st card.)
P(3rd card does not match the first 2 cards) = 8/10. (Of the 10 remaining cards, any but the mates of the first 2 cards.)
P(4th card does not match the first 3 cards) = 6/9. (Of the 9 remaining cards, any but the mates of the first 3 cards.)
Since we want all of these events to happen, we MULTIPLY the fractions:
10/11 * 8/10 * 6/9 = 16/33.
Thus:
P(at least 1 pair) = 1  16/33 = 17/33.
The correct answer is C.
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The chance that Bill finds at least one pair of cards that have the same value is nothing but the probability that there is no, at least, one pair of cards that have same value.
P(at least one pair of cards that have the same value)+P(no pair of cards that have same value)=1
=>P(at least one pair of cards that have the same value)=1P(no pair of cards that have same value)
P(no pair of cards that have same value)=(12/12)*(10/11)*(8/10)*(6/9)
12/12 probability of drawing any card
10/11 probability of drawing the card i.e., different from previous drawn
8/10 probability of drawing the card i.e., different from previous two draws
6/9 probability of drawing the card i.e., different from previous three draws
P(at least one pair of cards that have the same value)=1P(no pair of cards that have same value)
=>1(12/12)*(10/11)*(8/10)*(6/9)
=>1(16/33)
=>17/33
P(at least one pair of cards that have the same value)=17/33
P(at least one pair of cards that have the same value)+P(no pair of cards that have same value)=1
=>P(at least one pair of cards that have the same value)=1P(no pair of cards that have same value)
P(no pair of cards that have same value)=(12/12)*(10/11)*(8/10)*(6/9)
12/12 probability of drawing any card
10/11 probability of drawing the card i.e., different from previous drawn
8/10 probability of drawing the card i.e., different from previous two draws
6/9 probability of drawing the card i.e., different from previous three draws
P(at least one pair of cards that have the same value)=1P(no pair of cards that have same value)
=>1(12/12)*(10/11)*(8/10)*(6/9)
=>1(16/33)
=>17/33
P(at least one pair of cards that have the same value)=17/33