## MGMAT CAT Probability and Combinatronics

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### MGMAT CAT Probability and Combinatronics

by bubbliiiiiiii » Sun Sep 06, 2015 10:16 pm
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?
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by [email protected] Revolution » Mon Sep 07, 2015 12:22 am
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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by [email protected] Revolution » Mon Sep 07, 2015 1:16 am
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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by [email protected] » Mon Sep 07, 2015 5:36 am
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
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
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by [email protected] » Mon Sep 07, 2015 5:36 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, 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
We can also solve the question using counting techniques.
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/gmat-counting?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 1-card we have 2 suits from which to choose.
For the 2-card we have 2 suits from which to choose.
For the 4-card we have 2 suits from which to choose.
For the 5-card 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,
Brent
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by [email protected] » Tue Sep 08, 2015 2:31 am
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 non-pair)
= (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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by bubbliiiiiiii » Tue Sep 08, 2015 2:50 am
Hi All,

Thanks for responses having alternate approaches.

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!

Matt/Brent/Max - Thanks again for helping with alternate solutions. Can you also help me evaluate my approach?
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by [email protected] » Tue Sep 08, 2015 3:04 am
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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by [email protected] » Tue Sep 08, 2015 3:07 am
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. > <

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by bubbliiiiiiii » Tue Sep 08, 2015 3:08 am
Hi Matt,

Thanks for the correction.

Mistake I did was to divide it by 4! rather than 4.

Regards,
Pranay
Regards,

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by bubbliiiiiiii » Tue Sep 08, 2015 3:10 am
[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

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by jain2016 » Sat Mar 26, 2016 11:14 pm
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

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.

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by GMATGuruNY » Sun Mar 27, 2016 2:32 am
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(at least 1 pair) = 1 - P(no pairs).

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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by mallireddy » Mon Mar 28, 2016 1:42 am
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)=1-P(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)=1-P(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

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