Probability and Combination Problems
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 beatthegmat
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Found this great set of probability and combination problems. Rememberthese types of questions appear infrequently on the GMAT, so don't overemphasize them too much in your studies!
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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.
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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 beatthegmat
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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.
If the consensus seems to be that it sucks then I'll take it down.
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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.
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
Samir

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Thanks Eric for putting this up.
It has good variety of questions and it gains momentum after question 28.
It has good variety of questions and it gains momentum after question 28.

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This is an excellent resource.
I also found these problems on Scribd to be extremely difficult.
I also found these problems on Scribd to be extremely difficult.
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 Practice_Problem_Solving_Questions_1.pdf
 10 extremely difficult probability and number properties problems, all public domain, from Scribd.
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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).
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).
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?
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?
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 unattached vertex.
The calculation should be the complete graph on 21 vertices and then remove 18 vertices : 1/2*(21*18)18=171
any opinions ?
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 unattached vertex.
The calculation should be the complete graph on 21 vertices and then remove 18 vertices : 1/2*(21*18)18=171
any opinions ?
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?
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?
I think the equation they use in the solution actually does take that diagonal into account. Try it with a pentagon or hexagon.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 unattached vertex.
The calculation should be the complete graph on 21 vertices and then remove 18 vertices : 1/2*(21*18)18=171
any opinions ?
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?
 tbasebal24
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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
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