Does this Question look like GMAT Question?

[gmat740]

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. 8/3
b. 62/165
c. 17/33
d. 103/165
e. 25/33


I would definitely quote the source of this question after a few responses.

But here is my opinion about this question.
This question is an utter waste of time.
Just to give a brief idea what qualifies me to claim this.
I gave GMAT a month back and I scored Q50(94%) but I did not find any question of this type, may be GMAC knows that we are solving these questions under a time crunch.Apart from my Q50 in real GMAT,my GMAT prep scores for Quant is Q49(1st Gprep) and Q51(2nd Gprep)

I hope this does proves that I am quiet comfortable seeing tough question(given the adaptive algorithm of GMAT exam).But no matter how tough the Questions are,they are always framed keeping in mind the time constraint.

I want some sincere opinions before I jump to any conclusion.The above are my personal views regarding this question


Please express your reasoning before you vote

[Musiq]

I have been teaching the GMAT for 9+ years now.

In my opinion, this is a poor representation of what the GMAT is. The GMAT tests concepts of Perms, Combs etc...but not to the extent this questions is asking and definitely not in the uninspired , non-standardized language like the question you quoted uses.

[gmat740]

Thanks Musiq for your feed back. I would love to quote the source after few more responses, and I believe you would be surprise to hear about the source

[goelmohit2002]

Yes I think that this is a GMAT type problem...

Not a difficult one too if we understand the concept....

this is similar problem of selecting 2 couples from a pair of say 5 couples...

Exact same principle needs to applied...

Required Probability = 1 - None of the cards match.

[gmat740]

could you please solve this question?
I mean considering the time crunch

[goelmohit2002]

could you please solve this question?
I mean considering the time crunch

Divide the 12 cards in 6 groups...

(1,1), (2,2), (3,3),(4,4), (5,5), (6,6)

Selecting four different groups = 6C4
Out of these groups we have to select one card each so that none of the number match = 2C1 * 2C1 * 2C1 * 2C1

Thus total ways so that none of the cards match = 6C4 * 2C1 * 2C1 * 2C1 * 2C1

Total ways of selection of 4 cards = 12C4

Thus Probabilty when none match = say X = (6C4 * 2C1 * 2C1 * 2C1 * 2C1) / (12C4)

Required Probabilty when atleast one match = 1-X = 17/33 = C.

[gmat740]

Selecting four different groups = 6C4
Out of these groups we have to select one card each so that none of the number match = 2C1 * 2C1 * 2C1 * 2C1

is the question talking about 4 different groups or 4 different cards?

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.

[ankitns]

Yes. It did not take me that long to solve it. Probably under a minute.

A) Total possible combinations: (12 * 11 * 10 * 9)/4!
B) Total possible Combinations with no matches: (12 * 10 * 8 * 6)/4! [Note we are skipping by 2 here because there are 2 cards with same value]

Probability of no matches = B/A = 16/33.

Probability of atleast 1 match = 1 - 16/33 = 17/33.

When you claim that this was waste of time, could you please specify how long it took you? and more importantly what method/thought process did you use?

Cheers.

could you please solve this question?
I mean considering the time crunch

[goelmohit2002]

is the question talking about 4 different groups or 4 different cards?

[/quote]

We are finding four different numbered cards only by dividing the cards in different groups of same number and picking one card each from the group...so that none of the cards is same....i.e. 4 different cards....

[gmat740]

Yes. It did not take me that long to solve it. Probably under a minute.

A) Total possible combinations: (12 * 11 * 10 * 9)/4!
B) Total possible Combinations with no matches: (12 * 10 * 8 * 6)/4! [Note we are skipping by 2 here because there are 2 cards with same value]

Probability of no matches = B/A = 16/33.

Probability of atleast 1 match = 1 - 16/33 = 17/33.

When you claim that this was waste of time, could you please specify how long it took you? and more importantly what method/thought process did you use?

Cheers.

could you please solve this question?
I mean considering the time crunch

I am satisfied with goelmohit's method. I was also thinking on the same lines.
Where did I went wrong?
I was lost in the language.I have not been practicing probability questions which such non-standardized language (as musiq quoted above)

I have some questions with your method :Why are you taking combinations?
The question asks us to make selections. I believe both are different.

@mohit: Did you see this question for the first time, or you solved this question earlier?

Thanks Guys for discussion.

[goelmohit2002]

I had not solved the exact same question....but similar ones earlier.

[Musiq]

A GMAT problem is not defined by how long it takes to solve. That is but one criteria.

It should also meet one other criteria of Standardized Tests: Should be accesible to everyone. Hence the word Standardized. A Math major should not possess an unfailry indiscriminate advantage than say an Art Major. Though in theory this is true and rarely so in practice, the GMAT atleast tries to word the problems in such a way that the majority of test takers have an equally difficult/easy time understandign the question.

This is just my opinion, of course. Others may freely disagree.

[ankitns]

I wanna say that I have not seen this before, but I have solved a lot of counting/probability problems so it is very much possible that I have seen it before and just dont recollect where...

In terms of my approach, the question asks "What is the chance that ..." which to me translates as "What is the probability that...". Most of times I start with the basics for problems that I dont know a formula/trick for...so in this case since it asks for probability of something...we can start with Probability P(A) = (total favorable outcomes)/(total possible outcomes).
I remember reading somewhere that whenever the question asks for probablity of "atleast" something happening...the best way to approach the problem is to use P(A) = 1 - P(B).

So in this case P(B) = (total NOT favorable outcomes)/(total possible outcomes)....So then its just the matter of couting the outcomes...this is the approach i used to solve this...

Do you or anyone else for that matter sees something wrong with this approach? I retaking the GMAT soon and would love to know whether I am not on the right track or not.


In terms of the language of this problem...i do not have have much of an opinion on it. The language used in this problem may or may not be appropriate for GMAT...but in terms of difficulty/conceptual test I do think that this can be a GMAT problem...



Appreciate the feedback.

Thanks.

Yes. It did not take me that long to solve it. Probably under a minute.

A) Total possible combinations: (12 * 11 * 10 * 9)/4!
B) Total possible Combinations with no matches: (12 * 10 * 8 * 6)/4! [Note we are skipping by 2 here because there are 2 cards with same value]

Probability of no matches = B/A = 16/33.

Probability of atleast 1 match = 1 - 16/33 = 17/33.

When you claim that this was waste of time, could you please specify how long it took you? and more importantly what method/thought process did you use?

Cheers.

could you please solve this question?
I mean considering the time crunch

I am satisfied with goelmohit's method. I was also thinking on the same lines.
Where did I went wrong?
I was lost in the language.I have not been practicing probability questions which such non-standardized language (as musiq quoted above)

I have some questions with your method :Why are you taking combinations?
The question asks us to make selections. I believe both are different.

@mohit: Did you see this question for the first time, or you solved this question earlier?

Thanks Guys for discussion.

[tom4lax]

Antkins,

Would you mind possibly elaborating on how you got to Part B?

I can do this problem by thinking intuitively about what its asking:

1*10/11*8/10*6/9* = 16/33 answer is 17/33

However I don't want to rely on this method come test time if I am unable to figure it out.

[ankitns]

Part B --> (12 * 10 * 8 * 6)/4!

1) For the first selection we can select any of the 12 cards. Hence we get 12.
2) Now we are looking to not repeat the value in second selection. Now each time we select a card, it reduces the available choices by the number of duplicates (number of cards with the same value). In this case it is 2. Hence we get 12-2 ------> 10
3) Similarly, for the third selection we get 10-2 ----> 8
4) Finally, for the last selection we get 8-2 ----> 6


Hence we get 12 * 10 * 8 * 6.

Now we divide by 4! because there are 4! ways to arrange the 4 selection....ABCD is same as ACDB is same as DCBA etc...

**Note: if there were 3 cards for each value we would get (12 * 9 * 6 * 3)/4!..as we would subtract by 3 for each selection.


Hope this helps. Lemme know if you still have any questions.

Thanks.



3) Similarly we have

Antkins,

Would you mind possibly elaborating on how you got to Part B?

I can do this problem by thinking intuitively about what its asking:

1*10/11*8/10*6/9* = 16/33 answer is 17/33

However I don't want to rely on this method come test time if I am unable to figure it out.