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by vertigo05 » Tue Feb 23, 2010 7:32 am
Thanx, Sanju09 and Ron.. Very well explained.
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by analyst218 » Wed Feb 24, 2010 10:57 am
well a easier way , IMO , is to

just 1 - [getting non-pair/ total number]
= 1 - [12C4 / 12C4]

however the numerator has to be 12x10x8x6/4! since say you pick 1 for the first card,
you have to eliminate the other 1 so you have 10, same with the 2nd 3rd cards..

so 1-[12x10x8x6/4!]/[12C4] = 1 - 240/495 = 17/33
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by lunarpower » Thu Feb 25, 2010 2:36 am
analyst218 wrote:well a easier way , IMO , is to

just 1 - [getting non-pair/ total number]
= 1 - [12C4 / 12C4]

however the numerator has to be 12x10x8x6/4! since say you pick 1 for the first card,
you have to eliminate the other 1 so you have 10, same with the 2nd 3rd cards..

so 1-[12x10x8x6/4!]/[12C4] = 1 - 240/495 = 17/33
well, that certainly works, but i wouldn't call it "easier". look at the enormous size of the numbers in your calculations, vs. the tiny size of the numbers in the other posters' calculations.
(in the other version, the largest number in the entire workup is the actual answer to the problem, 17/33)

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also, this is really important:

WHEN YOU DO ARITHMETIC:
SIMPLIFY AS MUCH AS POSSIBLE BEFORE YOU WORK OUT COMPUTATIONS!

your calculation above says "240/495". there's no need to deal with numbers this big, and, unless you are absolutely lightning fast at calculations, there's no doubt that you wasted some time getting to that point (...and will waste even more time boiling those numbers back down to normal size).

here's what you do:
(the color-coded numbers CANCEL OUT)

[12x10x8x6/4!]/[12C4]

= (12 x 10 x 8 x 6 / 4 x 3 x 2 x 1) (4 x 3 x 2 x 1 / 12 x 11 x 10 x 9)

= (10 x 8 x 6 / 2 x 1) (1 x 2 / 11 x 10 x 9)

= (8 x 6) (1 / 11 x 9)

= (8 x 2) (1 / 11 x 3)

= 16/33

no need to deal with such awfully huge numbers.

i still submit that this is much, much more complicated than the sequential-multiplication solution submitted by the other posters.
but the most important thing here - it doesn't matter whether it's easier; it just matters whether you can come up with it right away.
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by analyst218 » Thu Feb 25, 2010 6:57 am
yes that's why I said IMO..
to me, multipying everything is easier.
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by gmatpill » Tue Jul 13, 2010 3:03 pm

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

Step 1:
Recognize the keyword "AT LEAST"

This means this is a "1 - probability()" question...this is the default strategy when a question is asking the probability of at least X happening. It translates into 1 - probability(that event NOT happening)

Step 2:
Now you need to find the probability of not getting a pair for each of the 4 cards. Recognize that as you go through each card, the number of available cards from the remaining deck is constantly changing. Originally you have 12 cards, but then after you put down 1 card, u have 11 cards left.


Step 3:
1st card = any card = prob of 1
2nd card = any card except the value of first card = 10 out of 11 choices
3rd card = any card except first card and except second card = 8 out of 10 choices
(notice we must exclude the 1st card and the 2nd card--but also the corresponding pairs for each of these cards.
4th card = any card except first/second/third card (and their corresponding pair card) = 6 out of 9 choices

Now multiply these probabilities so you know what the probability of NOT getting any pairs for all 4 iterations.
10/11 * 8/10* 6/9 = 480 / 990

1 - 480/990 = 510/990 = 17/33

Answer choice is C (17/33)



I don't recommend using the nCr formula here, but if you want to use nCr's, then:

= 1 - prob(choosing all different cards
= 1 - prob(choosing 4 different cards out of 6) * (possibilities for first card to be of two suits, for second card to be of two suits, for third card to be of two suits, and for fourth card to be of two suits)
= 1 - (6c4 * 2 * 2 * 2 * 2) / (12c4)
= 1 - (15 * 16) / (( 12 * 11 * 10 * 9 )/(4*3*2*1))
= 1 - 240 / 495
= 255/495

= 51/ 99
= 17 / 33

But again, this is not the most efficient way to think through this question. The best way is to count according to the method I discussed first above.
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by lakshmi86 » Wed Aug 25, 2010 11:56 pm
logitech wrote:1 2 3 4 5 6
1 2 3 4 5 6

So we can choose 12x10x8x6 ways but we need to divide it by 4!

= 240 ways ( NO PAIR )

C(12:4) =495

So : (495-240)/495
But combinations itself means that the order does not matter right? why divide by 4! then?? can u please enlighten me on this?? if we choose 2 members from a team of 5 we say it there are 5C2 ways of doing it and we do not divide by 2! factorial. PLease help?
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by sanju09 » Thu Aug 26, 2010 12:31 am
lakshmi86 wrote:
logitech wrote:1 2 3 4 5 6
1 2 3 4 5 6

So we can choose 12x10x8x6 ways but we need to divide it by 4!

= 240 ways ( NO PAIR )

C(12:4) =495

So : (495-240)/495
But combinations itself means that the order does not matter right? why divide by 4! then?? can u please enlighten me on this?? if we choose 2 members from a team of 5 we say it there are 5C2 ways of doing it and we do not divide by 2! factorial. PLease help?
That's right! But 5C2 is not 5P2 at the same time. What do we do when we need to get nC2 from a given nP2?
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by lunarpower » Thu Aug 26, 2010 2:01 am
lakshmi86 wrote:
logitech wrote:1 2 3 4 5 6
1 2 3 4 5 6

So we can choose 12x10x8x6 ways but we need to divide it by 4!

= 240 ways ( NO PAIR )

C(12:4) =495

So : (495-240)/495
But combinations itself means that the order does not matter right? why divide by 4! then?? can u please enlighten me on this?? if we choose 2 members from a team of 5 we say it there are 5C2 ways of doing it and we do not divide by 2! factorial. PLease help?
yeah, but the poster here is not using a combination formula. he's using consecutive multiplication, which always implies that the positions for which items are being chosen are distinguishable (i.e., that "order matters").
therefore, if you are using consecutive multiplication and the positions for which you are choosing items are indistinguishable (i.e., "order doesn't matter"), then you must divide by the factorial(s) of the number(s) of indistinguishable positions.

you can see much more about the method of consecutive multiplication (which i call the "slot method") in the video recording for december 3, 2009, at the following link:
https://www.manhattangmat.com/thursdays-with-ron.cfm
Ron has been teaching various standardized tests for 20 years.

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by pkonduri » Mon Jun 27, 2011 6:43 am
I am replying to a dormant post, I have some issue with my approach.Someone kindly help:

Actually one way to solve this problem is calculate the complement of what is asked and subtract from 1 i.e.,{1-p} . I did no get this idea while taking the test and tried to solve this question by calculating the probability for atleast one pair that is (one pair + two different cards) or two pairs.

A) Probability of two pairs:
first chance:12/12
Second:1/11
Third:10/10
Fourth:1/9
So (12/12)*(1/11)*(10/10)*(1/9) = 1/99

A) Probability of one pair and the other cards being different:
first chance:12/12
Second:1/11
Third:10/10
Fourth:8/9

it will come out to 8/99

Finally (A) + (B) is coming out to be 1/11

It is evident that I am missing something , its a shame that i am not able to figure it out after looking into so many discussions that have happened. can someone help me out.Please!
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by Ian Stewart » Mon Jun 27, 2011 10:29 am
pkonduri wrote:
It is evident that I am missing something , its a shame that i am not able to figure it out after looking into so many discussions that have happened. can someone help me out.Please!
The issue is that you're not counting all of the ways you might get exactly one pair, or two pairs. When you calculate:
pkonduri wrote:
A) Probability of two pairs:
first chance:12/12
Second:1/11
Third:10/10
Fourth:1/9
So (12/12)*(1/11)*(10/10)*(1/9) = 1/99
you are assuming that the 1st and 2nd cards are pairs, and that the 3rd and 4th cards are paired. There are other possibilities - it might be that the 1st and 3rd cards are paired, and the 2nd and 4th cards are paired, for example.

Because you aren't considering all of the different orders in which pairs might appear, you're missing several cases and your answer is too low because of that.

I'd add that this question is really *designed* to be solved by finding the probability you do NOT get the result in question, and subtracting that probability from 1. If you do try to do this problem directly, it won't be quick and it won't be easy.
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by pkonduri » Mon Jun 27, 2011 11:09 am
Thanks Ian for the explanation! yes got the problem. Also its true that 1-P method is a lot easier than what i tried. Thanks Again!
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by lunarpower » Tue Jun 28, 2011 3:07 am
what ian said.

by the way, for much more on these sorts of exercises with probability, check out the FEBRUARY 17, 2011 study hall at the following link:
https://www.manhattangmat.com/thursdays-with-ron.cfm
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by LalaB » Sat Jan 28, 2012 5:05 am
my approach-

total=12C4

first select 4 pairs out of six , and then get 1 card from every pair. i.e.

4C6*2C1*2C1*2C1*2C1/12C/4=16/33

1-16/33=17/33
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by tomada » Sat Jan 28, 2012 11:51 am
Shulapa, I like the way you solved this. I think it's the most purely intuitive of the possible solutions.

shulapa wrote:Hi,
I got to the same answer as Brent but through a different method. Tell me whether this works for you guys.

It is easier to find out the chances of getting 4 card without any couple of them with the same value. This can be deducted from 1 to get the probability of at least one couple.

Card 1: can be any number at any suit. So we can chose 12 out of 12.
Card 2: can be any card from the remaining 11 besides the card, which we already hold its number. So it is 10 out of 11.
Card 3: can be any card from the remaining 10 besides the 2 cards, which we already hold the numbers. So it is 8 out of 10.
Card 4: the same idea, only now we have to deduct 3 invalid cards. So it is 6 out of 9.

Combining the option gives us:
Probability of no couples = 1*(10/11)*(8/10)*(6/9) = 16/33.
Now we can say that:
Probability of at least one couple = 1 - 16/33 = 17/33
I'm really old, but I'll never be too old to become more educated.
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by tomada » Sat Jan 28, 2012 11:53 am
Shulapa, I like the way you solved this. I think it's the most purely intuitive of the possible solutions.

shulapa wrote:Hi,
I got to the same answer as Brent but through a different method. Tell me whether this works for you guys.

It is easier to find out the chances of getting 4 card without any couple of them with the same value. This can be deducted from 1 to get the probability of at least one couple.

Card 1: can be any number at any suit. So we can chose 12 out of 12.
Card 2: can be any card from the remaining 11 besides the card, which we already hold its number. So it is 10 out of 11.
Card 3: can be any card from the remaining 10 besides the 2 cards, which we already hold the numbers. So it is 8 out of 10.
Card 4: the same idea, only now we have to deduct 3 invalid cards. So it is 6 out of 9.

Combining the option gives us:
Probability of no couples = 1*(10/11)*(8/10)*(6/9) = 16/33.
Now we can say that:
Probability of at least one couple = 1 - 16/33 = 17/33
I'm really old, but I'll never be too old to become more educated.
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