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Probability and Combination Problems

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Probability and Combination Problems

by beatthegmat » Mon Oct 22, 2007 1:04 pm
Found this great set of probability and combination problems. Remember--these types of questions appear infrequently on the GMAT, so don't over-emphasize them too much in your studies!
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Probability and Combinations.doc
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by thunderdogg » Mon Nov 26, 2007 8:14 pm
Question 6 from this problem set:

6. A credit card number has 5 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

a) 27.
b) 36.
c) 72.
d) 112.
e) 422.



---
So um... how does a 5 digit credit card have a 6th digit that is used to determine the 5th digit. This problem set might be a waste of time. ONly on problem 6 and the questions are clearly not very good.
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by beatthegmat » Mon Nov 26, 2007 9:25 pm
Thanks for this feedback, thunderdogg. Anyone else have opinions as to whether this problem set is useful?

If the consensus seems to be that it sucks then I'll take it down.
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by samirpandeyit62 » Mon Nov 26, 2007 9:32 pm
Hi Eric,
I think its a good set for P&C coz it has so many different types of P&C problems, so it would be really good for practise & to learn the general approach to solve diff kinds of P&C probs, I've not seen each & every soln though, some may be wrong, but the same can be discussed.

Thanks for posting the doc.
Regards
Samir
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by tripathimani » Fri Dec 26, 2008 4:09 pm
Thanks Eric for putting this up.

It has good variety of questions and it gains momentum after question 28.
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by UCLA97grad » Mon Sep 07, 2009 1:44 pm
This is an excellent resource.

I also found these problems on Scribd to be extremely difficult.
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10 extremely difficult probability and number properties problems, all public domain, from Scribd.
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by Alex Lbn » Wed Jul 28, 2010 9:28 am
Original post: thunderdogg
Posted: Mon Nov 26, 2007 10:14 pm
Question 6 from this problem set:

6. A credit card number has 5 digits (between 1 to 9). The first two digits are 12 in that order, the third digit is bigger than 6, the forth is divisible by 3 and the fifth digit is 3 times the sixth. How many different credit card numbers exist?

a) 27.
b) 36.
c) 72.
d) 112.
e) 422.



---
So um... how does a 5 digit credit card have a 6th digit that is used to determine the 5th digit. This problem set might be a waste of time. ONly on problem 6 and the questions are clearly not very good.

This problem actually makes sense. We have 5 slots which we need to fill with different numbers. And we are given that the first and the second are 1 & 2 respectively.
_1_ _2_ __ __ __

The third digit is bigger than 6, therefore it could be either 7 or 8 or 9, since the range is [1:9].
The fourth digit is divisible by 3, therefore it could be either 3 or 6 or 9.
The fifth digit is 3 times the sixth, therefore it's multiple of 3, so in a given range it could be either 3 or 6 or 9. By the way, the sixth digit could be either 1 or 2 or 3, since the product of 5th and 6th can't be more than 9. So the reference to the 6th digit in the question is absolutely sensible.
From this point, it looks like a simple combinatorics problem. We have 3 options in each of three decisions. So to calculate possible numbers we need to multiply 3 by 3 by 3. So the answer is 27 (A).
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by boysangur » Sun Aug 15, 2010 12:25 pm
Thanks a LOT for posting this. I'm going through Kaplan right now and these types of problems are kicking my butt. I could never fully wrap my head around these problems, I don't know why.

More on topic: has anyone looked at #24? In the answers, it says that the probability of choosing an even number is 1/2, when we have 8 digits available (0 through 9, without 1 or 4). How is the probability 1/2 though? Aren't there 3 even numbers (2, 6 and 8) and 4 odd numbers (3, 5, 7, and 9)? Are they considering zero to be an even number? Both Kaplan and Princeton Review are claiming it isn't. Is the test wrong? Are the books? Am I crazy?
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by eladoren » Wed Oct 06, 2010 12:34 pm
regarding question #32
The answer 170 is not correct in my opinion. it should be 171
the suggested solution is not considering the diagonal between the two neighbors of the un-attached vertex.
The calculation should be the complete graph on 21 vertices and then remove 18 vertices : 1/2*(21*18)-18=171
any opinions ?
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by K23 » Mon Oct 11, 2010 8:53 am
can anybody explain the problem 27) in how many ways can you sit 8 people on the bench if 3 of them must sit together?
thanks.
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by K23 » Mon Oct 11, 2010 8:54 am
UCLA97grad wrote:This is an excellent resource.

I also found these problems on Scribd to be extremely difficult.
hi,
this link is not working . can please post it again
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by ga1ex » Fri Aug 17, 2012 10:39 pm
This problem set is super helpful - thanks for posting.

I think I've found some more incorrect answers:

#37. The answer doesn't take into account the possibility of the second and third digits being the same. I think the right answer is C.

#41. The answer takes into account the possibility of the first digit being 0, and excludes that as an option, but it doesn't do the same for the possibility of the first and second digits being 0, or first second and third, or all four. I'm not sure what the right answer is, but I think it's somewhere around 3515.

Am I missing something?
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by ga1ex » Fri Aug 17, 2012 11:35 pm
eladoren wrote:regarding question #32
The answer 170 is not correct in my opinion. it should be 171
the suggested solution is not considering the diagonal between the two neighbors of the un-attached vertex.
The calculation should be the complete graph on 21 vertices and then remove 18 vertices : 1/2*(21*18)-18=171
any opinions ?
I think the equation they use in the solution actually does take that diagonal into account. Try it with a pentagon or hexagon.
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by ga1ex » Sun Aug 19, 2012 9:17 pm
Can anyone help me with #47? I understand why the answer key's answer is correct, but my approach was to figure out the probability that one secretary gets no reports (16/27) and the probability that two secretaries get no reports (1/27) and subtract those probabilities from 1, and the result is different (10/27). Can someone please explain why my approach doesn't work?
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by tbasebal24 » Tue Oct 16, 2012 10:59 am
Can someone explain this one for me... the answer it gives is 14/15 but that seems way to high of a possibility for an answer.

52. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

A. 21/50
B. 3/13
C. 47/50
D. 14/15
E. 1/5
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