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Probability MGMAT

by voodoo_child » Fri Jun 15, 2012 6:25 am
You have to choose 2 cards from a standard 52 deck card. What is the probability that you find a pair? (pair = same number or rank)

a) 1/26
b) 1/17
c) 1/13
d) 4/51
e) 1/4

OA -B

I was able to arrive at the correct answer using two methods. However, initially I got stuck with Method3 and Method4. Can someone please guide me esp. Method3 and Method4?



Method 1 : Using permutations:

First card => 52/52
Second card => 3/51

Probability = 3/51= 1/17

Method 2 : Combinations:

There are 13C1 * 4C2 pairs = 13*6 pairs.
Total number of combinations to choose two cards = 52C2

PRobability = (13*6)C1/52C2 = 13*6*2/52*51 = 1/17

Method 3: (This is where I need help)

Let's think of 13X4 cards as matrix.

For instance,

[HEART SPADES DIAMOND CLUBS]
[ 1 1 1 1 ]
[ 2 2 2 2 ]
[ 3 3 3 3 ]
AND SO ON........


One can choose any one card out of 13 possible unique numbers/ranks (i.e. rows in the matrix). (I am ignoring difference in face card for now) => 13C1
Now, corresponding to each of these numbers, there are three other same number/rank card but with a different face. (i.e. column in the matrix) => second card can be chosen in 3C1 ways

Therefore, total number of combinations = 13*3/52C2 = 13*6/52*51 = 1/34. Incorrect.

Method 4:
I thought to myself : May be there is an issue with choosing *the first* card (i.e. considering all elements in the matrix). What if I choose the first card in : 52C1 number of ways
The second card can be chosen in only : 3C1 ways because there are only three other cards left with the same rank/number but different face.

Probability = 52 * 3 / 52C2 = 3*6/51 = 6/17...What????

I am stuck in black hole esp. with Method 3 and Method 4.

Can anyone please help me?

Thanks
Voodoo
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by Jim@StratusPrep » Fri Jun 15, 2012 6:36 am
your logic is wrong on the last two. Even though only 3 cards satisfy the pair you are not choosing from 3 cards you are choosing from 51. Even your logic in number 2 is wrong (the denominator should be 52*51/2) - the math doesn't work out to 1/17.

Stick with the first way!
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by voodoo_child » Fri Jun 15, 2012 6:57 am
Jim@StratusPrep wrote:your logic is wrong on the last two. Even though only 3 cards satisfy the pair you are not choosing from 3 cards you are choosing from 51. Even your logic in number 2 is wrong (the denominator should be 52*51/2) - the math doesn't work out to 1/17.

Stick with the first way!
Jim - I am sorry to say but the math in Method 2 is indeed correct. ( I am not sure about the method. Hence, I have requested Mike @ Magoosh to comment on it. I really love his explanations and his style.)
Here's step-by-step Method2:
13C1 * 4c2 = 13*6.
52C2 = 52*51/2

Therefore (13*6)/(52*51/2) = 13*6*2/52*51 = 13*4*3/52*51 = 1/17! :)

Secondly, I am sorry, but there is no "the method" in Maths. From what I have been taught, if one's understanding is correct, *all* methods must lead to the correct answer. To give you a perspective, Ramanujan, a renowned Indian mathematician, failed class 10 exam because he used to solve one question using at least 12 methods. Not that I am a mathematician, or trying to be one on the GMAT, but solving the same problem with different methods helps me to increase my understanding and gain speed. It's fun.

I am sorry, but can you elaborate on why Method3 and Method4 are incorrect?

(13*6)C1/52C2 = 13*6*2/52*51 = 1/17
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by gmat_and_me » Fri Jun 15, 2012 7:31 am
Your first solution is very nice. I can not make sense of
2 and 3. 4 you are part right. The mistake that you made in
4 is that you used combination in denominator while you did
not use combination in numerator. Look at it this way. First
card you can select from any of the 52 cards. But the second
card can only be taken 3 more ways. So that indeed is 52 * 3.
But that also means that you count 1spade, 1diamond and 1diamond,
1spade (for eg:) as 2 separate things. But, in denominator you
took out duplicates by doing 52C2. That is wrong.

You can say,

all possibilities of selecting 2 cards = 52 * 51
# of possibilities of selecting two cards of same rank = 13 * 4 * 3 (or 52 * 3)
whichever way you want to look at.

HTH
voodoo_child wrote:You have to choose 2 cards from a standard 52 deck card. What is the probability that you find a pair? (pair = same number or rank)

a) 1/26
b) 1/17
c) 1/13
d) 4/51
e) 1/4

OA -B

I was able to arrive at the correct answer using two methods. However, initially I got stuck with Method3 and Method4. Can someone please guide me esp. Method3 and Method4?



Method 1 : Using permutations:

First card => 52/52
Second card => 3/51

Probability = 3/51= 1/17

Method 2 : Combinations:

There are 13C1 * 4C2 pairs = 13*6 pairs.
Total number of combinations to choose two cards = 52C2

PRobability = (13*6)C1/52C2 = 13*6*2/52*51 = 1/17

Method 3: (This is where I need help)

Let's think of 13X4 cards as matrix.

For instance,

[HEART SPADES DIAMOND CLUBS]
[ 1 1 1 1 ]
[ 2 2 2 2 ]
[ 3 3 3 3 ]
AND SO ON........


One can choose any one card out of 13 possible unique numbers/ranks (i.e. rows in the matrix). (I am ignoring difference in face card for now) => 13C1
Now, corresponding to each of these numbers, there are three other same number/rank card but with a different face. (i.e. column in the matrix) => second card can be chosen in 3C1 ways

Therefore, total number of combinations = 13*3/52C2 = 13*6/52*51 = 1/34. Incorrect.

Method 4:
I thought to myself : May be there is an issue with choosing *the first* card (i.e. considering all elements in the matrix). What if I choose the first card in : 52C1 number of ways
The second card can be chosen in only : 3C1 ways because there are only three other cards left with the same rank/number but different face.

Probability = 52 * 3 / 52C2 = 3*6/51 = 6/17...What????

I am stuck in black hole esp. with Method 3 and Method 4.

Can anyone please help me?

Thanks
Voodoo
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by voodoo_child » Fri Jun 15, 2012 8:09 am
gmat_and_me wrote:Your first solution is very nice. I can not make sense of
2 and 3.
In method 2/ method 3 are the same except for the fact that I did a careless mistake.

The objective of method2/3 was to first find *all* possible pairs from the deck. All possible pairs = 13C1 * 3C1 because
the first card can be chosen in 1 out of 13 ways. Now there are only 3 same-numbered cards left. HEnce, number of combinations = 13C1 * 3C1 = 39

Therefore, probability = 1 pair out of 39/ 2 cards out of 52 = 39C1/52C2 = 1/34. Incorrect.

Can you please help me? I love this problem.....

I will also analyze your commetns about method4. I have a sense of where I am going wrong. But let me try to absorb it....
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by Mike@Magoosh » Fri Jun 15, 2012 9:39 am
Hello voodoo_child,

Once again, I am happy to answer your questions. :)

First of all, I think we all agree, Method One is perfect, an efficient route to the answer. Incidentally, you call that "using permutations", but I would call that "probability methods".

Your Method Two is correct, but not necessarily as efficient as Method One, at least in this instance. I like that you understand that approach --- count the total number of two-of-a-kind pairs, and divided by the total number of two-card combinations. Very clever.

On Method Three ---
What's bizarre about this --- you start with, "One can choose any one card out of 13 possible unique numbers/ranks" --- but picking just the row (i.e. the number/rank) is not the same as picking an individual card. 13*3 = 39 is not particularly the number of anything. ----- This would be the correct approach if the question were something like "What's the probability that, on a two-card draw, you pick a two-of-a-kind pair in which one of the two cards is a heart?" That way, just the choice of the suit would be enough to determine a unique card, the heart.
If for each of the 13 rows, you pick the numbers of two-of-a-kind pairs you can find on that row, 4C2 = 6, then we are back to something that looks very much like Method Two.

For Method Four, the tricky thing is that you double-counted cards, and you also made an arithmetic mistake. First of all, following your logic, the numerical answer should be ---

52*3/(52C2) = 52*3*2/(52*51) = 3*2/51 = 2/17

That's twice the correct answer. Why is that twice the correct answer? Because you counted each pair twice.

When you count all the first card choices, one of them could be, say, the Seven of Clubs. Then you count the three cards that can match that, one of which will be the Seven of Diamonds. Well, somewhere else in the process of picking all 52 cards for the first card, the Seven of Diamonds would have to get picked first, and one of its three matches is the Seven of Clubs. That pair gets counted twice. In such a fashion, each pair gets counted twice, so you have to divide this answer by two to eliminate the redundancy, in which case this equals the correct answer.

Really, Method Three, when we correct it, is just Method Two in disguise, but you do have three genuine different ways to answer this question. Great work!

Let me know if you have any further questions.

Mike :)
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by voodoo_child » Sat Jun 16, 2012 6:36 am
Mike@Magoosh wrote:Hello voodoo_child,



On Method Three ---
What's bizarre about this --- you start with, "One can choose any one card out of 13 possible unique numbers/ranks" --- but picking just the row (i.e. the number/rank) is not the same as picking an individual card. 13*3 = 39 is not particularly the number of anything. ----- This would be the correct approach if the question were something like "What's the probability that, on a two-card draw, you pick a two-of-a-kind pair in which one of the two cards is a heart?" That way, just the choice of the suit would be enough to determine a unique card, the heart.
If for each of the 13 rows, you pick the numbers of two-of-a-kind pairs you can find on that row, 4C2 = 6, then we are back to something that looks very much like Method Two.
Mike, I want to thank you for answering my questions. Do you mind explaining the above idea again? I didn't quite get it. Why is it that 13C1 * 3c1 incorrect but 13C1 * 4c2 correct? I see your point that 13C1*3c1 implies drawing one card after the other in such a way that if the first card is say seven of Spades, the second card should be seven of the remaining three face cards.

I am not able to get the difference between the above concept and 4C2. The only thing I can think of is that 4C2 ignores WHAT FACE CARD is drawn in the first attempt.

Please help me :(

Thanks
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by Mike@Magoosh » Mon Jun 18, 2012 11:01 am
voodoo_child wrote: Mike, I want to thank you for answering my questions. Do you mind explaining the above idea again? I didn't quite get it. Why is it that 13C1 * 3c1 incorrect but 13C1 * 4c2 correct? I see your point that 13C1*3c1 implies drawing one card after the other in such a way that if the first card is say seven of Spades, the second card should be seven of the remaining three face cards.

I am not able to get the difference between the above concept and 4C2. The only thing I can think of is that 4C2 ignores WHAT FACE CARD is drawn in the first attempt.

Please help me :(

Thanks
Voodoo
Dear Voodoo Child,

As always, I am happy to answer your excellent questions. :)

Suppose we have the 52 cards of the deck in a matrix, as you proposed --- each of the four columns in a different suit, and each of the 13 rows is a different number/rank. We want to count the number of two-of-a-kind pairs.

As you described your Method Three approach, your 13C1 choice was a choice of one thing out of 13 things --- in other words, the choice of one row from the thirteen rows. That was not the choice of an individual card, because there is absolutely no possible choice of one-from-thirteen that would isolate a single card from the deck of 52 card. This is the thing that is so mindbogglingly subtle about counting problems --- if the thing you intend to count is not properly represented by the mathematical expression you have chosen, then in fact you wind up counting something completely different from what you intended.

So, you intended to pick a first card, but in writing 13C1, you didn't actually pick a single card. Instead, you picked an entire row of the matrix, and that row has four equal-rank cards in it. That's why your next step can't be 3C1 to pick a second card --- you can't pick a second card if you haven't already picked a first card! When the mathematical expression does not represent the first choice you intended to make, that often has the effect of making the the second choice nonsensical. If you had actually picked a first card. a unique individual card, as you intended to, then picking a second card of the same rank would make all the sense in the world. Instead, you inadvertently picked an entire row of the matrix--- say, the row with the four 3's in it ---- so picking "second card", a second 3, after you already picked all four 3's altogether at once --- that doesn't make sense. Do you see what I mean?

If your first selection is 13C1, that means your first selection picks a row of the matrix, not an individual card. Once we have the row from which our two-of-a-kind pair will be drawn, what would be the logical next step? Once we have the row, we have to pick two of the four cards in in that row ----- 4C2. Again, if you follows out this method --- number of two-of-a-kind pairs = (13C1)*(4C2), which is exactly what we had in Method Two. Method Three and Method Two differ in superficial "story", but not in underlying mathematical structure.

If you wanted your first choice to isolate a single card, a unique individual card, from the matrix, I think the only way to do that would be 52C1 --- a choice of one card from all 52. Then, the first choice would legitimately be a single unique card, so your second choice could be 3C1 --- then, the mathematical approach is identical to your Method One.

Basically, Methods One, Two, and Four are genuinely different approaches to this problem ---- it's wonderful that you found three valid different approaches. In Method Three, you were trying to create something different from these other three, but when we correct Method Three, it reduces to one of the other methods. Understand, that's not bad. I have a tremendous amount of experience with counting and probability, and I can't think of an additional approach that is meaningful different from the three approaches you presented here. You understand this problem very well.

Does all this make sense? I believe all your confusion in Method Three comes from the fact that what you intended to choose and what you actually chose were different, because the mathematical expression you wrote had different implications from what you intended.

Does all this make sense? Please let me know if you have any further questions.

Mike :-)
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by dhonu121 » Tue Jun 19, 2012 10:37 am
I am using the following approach and am not able to get any of the answer choices.
Number of ways to select two cards without replacement of the same group = 4C1*13C2.
Number of ways to select any two cards = 52C2.
Thus required = 4C1*13C2/52C2.

This is coming out to be 4/17 and the correct answer is 1/17.
What is wrong that I am doing ?
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by Mike@Magoosh » Tue Jun 19, 2012 12:13 pm
dhonu121 wrote:I am using the following approach and am not able to get any of the answer choices.
Number of ways to select two cards without replacement of the same group = 4C1*13C2.
Number of ways to select any two cards = 52C2.
Thus required = 4C1*13C2/52C2.

This is coming out to be 4/17 and the correct answer is 1/17.
What is wrong that I am doing ?
Hi, there. I'm happy to help. :)

Your first step doesn't make sense to me ----- 4C1 means you are going to picking one option from four possible options, and 13C2 means you are going to pick two options from 13 possible options. For example, if the question were --- what's the probability that you draw two cards from the deck and they both are the same suit? ---- that question would be answered by your approach --- the 4C1 choice would be the choice of one suit from among the four possible suits, and the 13C2 would be two cards within that suit.

The trouble is --- that doesn't constitute a pair, a two-of-a-kind pair, in a poker sense. (I don't know how familiar you are with poker.) A "pair" in poker are two cards of different suits but of the same number/rank. For example, the 8 of Diamonds and 8 of Hearts would be a pair; the Queen of Hearts and the Queen of Spades would be a pair. What you have calculated here, two cards of the same suit --- e.g. the Jake of Clubs and the four of Clubs --- that's not a "pair" in the poker sense, although it could be the beginning of a "flush" (another poker term).

It appears to me that you understand the math, but have trouble with the poker definitions of card combinations. I would suggest this article . . .
https://en.wikipedia.org/wiki/Poker_hands
... as an excellent introduction to the terminology, as well as an excellent presentation of the relevant mathematics.

Let me know if you have any further questions.

Mike :)
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by dhonu121 » Tue Jun 19, 2012 8:27 pm
Hi Mike,
If that is the definition of poker pair,
the what is wrong with the following approach to calculate the desired probability
Number of ways to draw two cards of same pair wthout replcement = 13C1*4C2.
Total number of ways = 52C2.
Prob = (13C1*4C2)/52C2
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by Mike@Magoosh » Wed Jun 20, 2012 10:49 am
dhonu121 wrote:Hi Mike,
If that is the definition of poker pair,
the what is wrong with the following approach to calculate the desired probability
Number of ways to draw two cards of same pair wthout replcement = 13C1*4C2.
Total number of ways = 52C2.
Prob = (13C1*4C2)/52C2
Dear dhonu121:
Nothing is wrong with that. :) That gives the correct answer of 1/17, and in fact, it's identical to voodoo child's "Method One" in the top post in this thread. Just so you see how the calculation simplifies ---

13C1 = 13

4C2 = (4*3*2*1)/[(2*1)(2*1)] = (4*3)/(2*1) = 6

52C2 = (52*51)/2

(13C1*4C2)/52C2 = (13)(6)/[(52*51)/2] = (2)(13)(6)/(52*51) = (2)(6)/(4*51)

= (6)/(2*51) = 3/51 = 1/17

Does all that make sense? Let me know if you have any further questions.

Mike :)
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by dhonu121 » Wed Jun 20, 2012 11:51 am
Got it.
Thanks.
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by gmattesttaker2 » Wed Jun 20, 2012 6:54 pm
Mike@Magoosh wrote:
dhonu121 wrote:Hi Mike,
If that is the definition of poker pair,
the what is wrong with the following approach to calculate the desired probability
Number of ways to draw two cards of same pair wthout replcement = 13C1*4C2.
Total number of ways = 52C2.
Prob = (13C1*4C2)/52C2
Dear dhonu121:
Nothing is wrong with that. :) That gives the correct answer of 1/17, and in fact, it's identical to voodoo child's "Method One" in the top post in this thread. Just so you see how the calculation simplifies ---

13C1 = 13

4C2 = (4*3*2*1)/[(2*1)(2*1)] = (4*3)/(2*1) = 6

52C2 = (52*51)/2

(13C1*4C2)/52C2 = (13)(6)/[(52*51)/2] = (2)(13)(6)/(52*51) = (2)(6)/(4*51)

= (6)/(2*51) = 3/51 = 1/17

Does all that make sense? Let me know if you have any further questions.

Mike :)
Hi Mike,

Sorry to bother you but I am still not clear with Vodoo Child's Method 2 explanation:

Method 2 : Combinations:

There are 13C1 * 4C2 pairs = 13*6 pairs.
Total number of combinations to choose two cards = 52C2

PRobability = (13*6)C1/52C2 = 13*6*2/52*51 = 1/17

I was trying to approach this as follows:

P(Event) = number of ways the event can occur/Total number of possible outcomes

13C1 is clear i.e. we are drawing 1 card out of 13 cards. However I am lost after that.

52C2 is clear i.e. 2 cards out of 52.

Can you please help explain this?


Also, for Method1 i.e. Permutations I am trying to understand as follows:

For First card, total choices are 52. So first card could be any of the 52 i.e.

52/52

So let us say, I draw King of Hearts. Hence, the second card should be either a King of Diamonds or
a King of Spades or a King of Clubs. So it has to be these 3 out of a possible 51 i.e.

3/51

Hence, number of ways I can get a pair = 52/52 * 3/51 = 1/17

Is this understanding correct?

Thanks for your valuable time and help.

Best Regards,
Sri
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by Mike@Magoosh » Thu Jun 21, 2012 8:33 am
gmattesttaker2 wrote:Hi Mike,

Sorry to bother you but I am still not clear with Vodoo Child's Method 2 explanation:

Method 2 : Combinations:
There are 13C1 * 4C2 pairs = 13*6 pairs.
Total number of combinations to choose two cards = 52C2
Probability = (13*6)C1/52C2 = 13*6*2/52*51 = 1/17
I was trying to approach this as follows:
P(Event) = number of ways the event can occur/Total number of possible outcomes
13C1 is clear i.e. we are drawing 1 card out of 13 cards. However I am lost after that.
52C2 is clear i.e. 2 cards out of 52.
Can you please help explain this?
Sri ---
I am more than happy to help with this. :)
I'm not sure how familiar you are with nCr notation and the idea of combinations. Here's a blogpost that covers the basic.
https://magoosh.com/gmat/2012/gmat-permu ... binations/

In Voodoo Child's Method One, we first count the number of two-of-a-kind pairs, poker pairs, in the deck. The cards of a pair must be the same number/rank --- i.e. both 3's, or both Jack's. There are 13 numbers/ranks in the deck, so first we count that. BTW, writing that as 13C1 is, I would argue, overkill of the C notation.

Next, for the four cards of any given rank, how many different two card pairs can we create? This is a valid use of the nCr notation. We have 4 choose 2, which is

4C2 = (4*3*2*1)/[(2*1)(2*1)] = (4*3)/(2*1) = 6

Again, if you are unfamiliar with the nCr formula, please read the blogpost at that link --- it will explain everything clearly.

So, the total number of pairs is the number of numbers/ranks (13) times the number of pairs within each rank (6). We'll just leave this as 13*6 --- there's absolutely no reason on Earth to multiply that out, because we know we are going to cancel factors in a fraction. Also, I know Voodoo Child wrote this as (13*6)C1, but that is just an insane use of C-notation --- Voodoo Child went a little C-happy in that write-up. The numerator of our fraction is just plain old 13*6.

Now, the denominator. We have 52 cards, and we want to choose 2, so that's 52C2, "52 choose 2". Here's the long way to calculate it.

52C2 = (52!)/[(2!)(52 - 2)!] = (52!)/[(2!)(50)!] = (52*51*50!)/(2*(50!)) = 52*51/2

The shortcut is to recognize that nC2 always equals n(n-1)/2. Anyway, that is our denominator.

Probability = (number of desired events)/(total number of events)
= (13*6)/[52*51/2] ----- dividing by a fraction means multiplying by its reciprocal

= (13*6)*(2/(52*51) ----- now, cancel the factor of 13 (because 52 = 4*13)

= (6*2)/(4*51) --- now, cancel a factor of 2

= (6)/(2*51) --- now, cancel another factor of 2

= (3)/(51) --- now, cancel a factor of 3

= 1/17
gmattesttaker2 wrote:Also, for Method 1 i.e. Permutations I am trying to understand as follows:

For First card, total choices are 52. So first card could be any of the 52 i.e.

52/52

So let us say, I draw King of Hearts. Hence, the second card should be either a King of Diamonds or
a King of Spades or a King of Clubs. So it has to be these 3 out of a possible 51 i.e.

3/51

Hence, number of ways I can get a pair = 52/52 * 3/51 = 1/17

Is this understanding correct?
Yes, that is 100% correct. I hope this is helpful. Let me know if you have any more questions.

Mike :)
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