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Re: Probability & Cominatorics Test #10 [#permalink]
06 Aug 2009, 12:27

I am little skepticle about the question. The question is silent about the order of outcome i.e. an outcome of (21 and 23) and (23 and 21). IMO, an outcome of (21 and 23) is the same as (23 and 21) because the order here does not matter.

Total numbers = 20, 21, 22, 23, 24,25, 26, 27, 28, & 29. 10 numbers 2 out of 10 = 10c2 +10 = 55

primes = 23 and 29 multiple of 3 = 21, 24, and 27

Getting a prime and a multiple of 3 = 2x3 = 6 The prob = 6/55 = 11%

bipolarbear wrote:

What is the probability that one of the two integers randomly selected from range 20-29 is prime and the other is a multiple of 3?

(The numbers are selected independently of each other, i.e. they can be equal)

(C) 2008 GMAT Club - Probability and Combinations#10

Re: Probability & Cominatorics Test #10 [#permalink]
06 Aug 2009, 13:56

possible primes - 23, 29 = 2 multiples of 3 - 21, 24, 27 = 3 total numbers = 10

possibility of getting prime and then multiple of 3 2/10 * 3/10 = 6/100

the possibility of getting multiple of 3 and then prim 3/10 * 2/10= 6/100

Answer 12/100

The best way to think about why you don't multiply it is because it will decrease the probability to 3.6/100, but we already know that the answer (6/100) is too specific! Order here does not matter and 6/100 gives the probability of a specific order.

Re: Probability & Cominatorics Test #10 [#permalink]
10 Aug 2009, 22:03

bipolarbear wrote:

What is the probability that one of the two integers randomly selected from range 20-29 is prime and the other is a multiple of 3? (The numbers are selected independently of each other, i.e. they can be equal)

(C) 2008 GMAT Club - Probability and Combinations#10

Re: Probability & Cominatorics Test #10 [#permalink]
08 Oct 2013, 22:07

I have seen the official solution, nice and elegant by the way:

3 = \frac{2}{10} \frac{3}{10} = 0.06 .

The probability that the first number is a multiple of 3 while the second is prime = \frac{3}{10} \frac{2}{10} = 0.06 .

The probability that one of the two integers is prime and the other is a multiple of 3 = 0.06 + 0.06 = 0.12

It is all clear.

But, I began to solve this thing using the combinations...I know the above is a faster approach for many people (including myself), but I want to finish the thing using combinations method, which should give exactly 12% the way probability method does.

Here is my thinking: 1) ways to pick 2 out 10 numbers - 10C2 = 45 We have a total of 2 of primes and we need to pick 1, so 2C1 = 2. We have a total of 3 multiples of 3 we need to pick 1, so 3C1 = 3

Re: Probability & Cominatorics Test #10 [#permalink]
09 Oct 2013, 00:43

Expert's post

obs23 wrote:

I have seen the official solution, nice and elegant by the way:

3 = \frac{2}{10} \frac{3}{10} = 0.06 .

The probability that the first number is a multiple of 3 while the second is prime = \frac{3}{10} \frac{2}{10} = 0.06 .

The probability that one of the two integers is prime and the other is a multiple of 3 = 0.06 + 0.06 = 0.12

It is all clear.

But, I began to solve this thing using the combinations...I know the above is a faster approach for many people (including myself), but I want to finish the thing using combinations method, which should give exactly 12% the way probability method does.

Here is my thinking: 1) ways to pick 2 out 10 numbers - 10C2 = 45 We have a total of 2 of primes and we need to pick 1, so 2C1 = 2. We have a total of 3 multiples of 3 we need to pick 1, so 3C1 = 3

Prime integers: 23 and 29. Multiples of 3: 21, 24, and 27.

The probability that the first number is prime while the second is a multiple of 3 = \frac{2}{10} \frac{3}{10} = 0.06 .

The probability that the first number is a multiple of 3 while the second is prime = \frac{3}{10} \frac{2}{10} = 0.06 .

The probability that one of the two integers is prime and the other is a multiple of 3 = 0.06 + 0.06 = 0.12 . The correct answer is B.

Why are they added together and not multiplied together?

Hi Bunuel,

I don't why are we considering the order here. Does it really matter whether a multiple of 3 is chosen first and then a prime??

It just says probability that one of the two integers randomly selected.. So if we get 21,23 or 23,21.. How does it matter. One of the integers is a multiple of 3 and other is prime in both the cases..

Re: Probability & Cominatorics Test #10 [#permalink]
09 Oct 2013, 23:02

Quote:

Let me ask you a question: does 10C2 give the following possible pairs: (20, 20), (21, 21), ..., (29, 29) ?

Apparently, it does not, but it should...That is a good question, why does not this formula include them? Or in other words, what does my 10C2 assume? I kind of see it is not correct and the subtle difference but I am afraid I am unable to articulate it to myself.

The way I see it now is that you have to think about this problem not in terms of one set of 10 numbers but somehow in terms of two different sets of identical numbers? Then it kind of makes sense with 10x10 / 2! - I still need a proper way to think about it, otherwise tomorrow I wont be able to solve a similar problem.

Either way, I also do not see clearly the problem with 10C2.

Looks like the good news is that I am correct on the second part: 2C1 * 3C1.

P.S. From what opera of combinatorics is this? I mean what is the concept behind this particular stuff? Kudos to the problem! _________________

There are times when I do not mind kudos...I do enjoy giving some for help

Re: Probability & Cominatorics Test #10 [#permalink]
10 Oct 2013, 01:40

1

This post received KUDOS

Expert's post

obs23 wrote:

Quote:

Let me ask you a question: does 10C2 give the following possible pairs: (20, 20), (21, 21), ..., (29, 29) ?

Apparently, it does not, but it should...That is a good question, why does not this formula include them? Or in other words, what does my 10C2 assume? I kind of see it is not correct and the subtle difference but I am afraid I am unable to articulate it to myself.

10C2 gives 2 distinct numbers out of 10, which is not our case. _________________

Re: Probability & Cominatorics Test #10 [#permalink]
10 Oct 2013, 06:05

Bunuel wrote:

obs23 wrote:

Quote:

Let me ask you a question: does 10C2 give the following possible pairs: (20, 20), (21, 21), ..., (29, 29) ?

Apparently, it does not, but it should...That is a good question, why does not this formula include them? Or in other words, what does my 10C2 assume? I kind of see it is not correct and the subtle difference but I am afraid I am unable to articulate it to myself.

10C2 gives 2 distinct numbers out of 10, which is not our case.

That makes sense - could you provide final explanation using combs here? Or is this the correct way then \frac{3C1*2C1}{(10*10)/2!} = \frac{6}{50}=0.12?We have here several posts with a total number of combs 55, and in excel it does seem to be the case as well, so I am still confused. _________________

There are times when I do not mind kudos...I do enjoy giving some for help

Re: Probability & Cominatorics Test #10 [#permalink]
10 Oct 2013, 06:27

Expert's post

obs23 wrote:

That makes sense - could you provide final explanation using combs here? Or is this the correct way then \frac{3C1*2C1}{(10*10)/2!} = \frac{6}{50}=0.12?We have here several posts with a total number of combs 55, and in excel it does seem to be the case as well, so I am still confused.

For some questions it's better to use probability approach. This is one of those. _________________

Re: Probability & Cominatorics Test #10 [#permalink]
11 Oct 2013, 09:55

Expert's post

obs23 wrote:

That makes sense - could you provide final explanation using combs here? Or is this the correct way then \frac{3C1*2C1}{(10*10)/2!} = \frac{6}{50}=0.12?We have here several posts with a total number of combs 55, and in excel it does seem to be the case as well, so I am still confused.

obs23 wrote:

Hello Mike! Hope you are all well. Was wondering if you could step in to explain how the problem can be solved using combinatorics? I understand it is not practical, but just for the fundamentals sake as well as piece of mind, I would like to know that approach even if it takes forever. Would really appreciate your input. Regards, Obs

Dear Obs, I got your private email and I am happy to help.

Notice, first of all, that GMAT TIGER, in the very second post in this thread, skillfully used some combinatorics quite correctly in his wonderful analysis of the question. What you want to see is already demonstrated on the page.

I would also say: something about what you are asking a question that is not well thought out. Let me explain. You say, you want to use combinatorics. Think about this. What is combinatorics? Combinatorics is a way of counting things that would be hard to list explicitly, such as combinations and permutations. If I have, say eight members of a group, and I want to pick a combination of three, it would be time-consuming to list all of them out --- we would just use 8C3. The only purpose of combinatorics is to count things. Well, once we have counted, by any means available, then we are done with combinatorics. It does not enhance your problem-solving if you make the counting process inherently more complicated by introducing combinatorics unnecessarily. The entire point of problem-solving is to make everything as clear & direct & efficient as possible. In many problems, there is more than one efficient approach to a problem, and that's great to see, but you are not enhancing your problem-solving instincts if you are finding more difficult or more time-consuming ways to solve a problem.

Notice, also, in your attempt to solve this question with combinatorics, you jumped immediately to a formula. If the first question you ask in approaching a GMAT math problem is, "What formula should I use?", then you choosing a wildly unsuccessful strategy, and the GMAT Quant will frustrate you time and time again. The GMAT Quant section is designed to punish users who are too formula-based in their approach.

Re: Probability & Cominatorics Test #10 [#permalink]
14 Oct 2013, 03:16

mikemcgarry wrote:

obs23 wrote:

That makes sense - could you provide final explanation using combs here? Or is this the correct way then \frac{3C1*2C1}{(10*10)/2!} = \frac{6}{50}=0.12?We have here several posts with a total number of combs 55, and in excel it does seem to be the case as well, so I am still confused.

obs23 wrote:

Hello Mike! Hope you are all well. Was wondering if you could step in to explain how the problem can be solved using combinatorics? I understand it is not practical, but just for the fundamentals sake as well as for the peace of mind, I would like to know that approach even if it takes forever. Would really appreciate your input. Regards, Obs

Dear Obs, I got your private email and I am happy to help.

Notice, first of all, that GMAT TIGER, in the very second post in this thread, skillfully used some combinatorics quite correctly in his wonderful analysis of the question. What you want to see is already demonstrated on the page.

I would also say: something about what you are asking a question that is not well thought out. Let me explain. You say, you want to use combinatorics. Think about this. What is combinatorics? Combinatorics is a way of counting things that would be hard to list explicitly, such as combinations and permutations. If I have, say eight members of a group, and I want to pick a combination of three, it would be time-consuming to list all of them out --- we would just use 8C3. The only purpose of combinatorics is to count things. Well, once we have counted, by any means available, then we are done with combinatorics. It does not enhance your problem-solving if you make the counting process inherently more complicated by introducing combinatorics unnecessarily. The entire point of problem-solving is to make everything as clear & direct & efficient as possible. In many problems, there is more than one efficient approach to a problem, and that's great to see, but you are not enhancing your problem-solving instincts if you are finding more difficult or more time-consuming ways to solve a problem.

Notice, also, in your attempt to solve this question with combinatorics, you jumped immediately to a formula. If the first question you ask in approaching a GMAT math problem is, "What formula should I use?", then you choosing a wildly unsuccessful strategy, and the GMAT Quant will frustrate you time and time again. The GMAT Quant section is designed to punish users who are too formula-based in their approach.

Does all this make sense? Mike

Mike - thanks a lot for your response - I really appreciate your input. Notice I do not go with the formulaic approach.

Quote:

Here is my thinking (order does not matter): 1) ways to pick 2 out 10 numbers - 10C2 = 45 We have a total of 2 of primes and we need to pick 1, so 2C1 = 2. We have a total of 3 multiples of 3 we need to pick 1, so 3C1 = 3

I am breaking the whole thing apart by "thinking it out loud"(we have already agreed with Bunuel 45 is wrong, it does not include 20,20, ...29,29 combinations) in an organized way and understandable way with as few assumptions as possible (this is how my brain works). Apparently, there is no disagreement about my thinking on the numerator part (we have 6 ways to choose a multiple of 3 and a prime), but it seems to me we can't get to the denominator part, using the same logic...I am bewildered, how come?! I do not think it is impossible. If 55 is correct, looks like there is no disagreement on that either and 10c2 +10 = 55, then 6 must be incorrect... or vice versa. 12% (probability based) is a hard number in my mind, it is proven above, and it is so darn logical. For any problem there must be a theoretical approach... and either way, all ways must have an identical result.

What I am concerned about at the end of the day, with all due respect to GMATTIGER and others agreeing with his solution, is that we have two different results. 10.91% and 12% (straight out, no rounding) - so I am puzzled, how is this possible (the way it stands right now, is that GMATTIGER is not correct, even though he used the method skillfully)?! GMATTIGER must be missing something, and if the answer choices contained 11% as an answer, he would simply be wrong. I just want to get to the same answer both ways and understand where he is incorrect (I just can't see what he is missing! [we are using probability theory that supposed to be correct!]) - and I mentioned above, I understand the second approach is more difficult one (the problem is I began to think in described above terms right away and by the time I realized I need turn other way, I have already spent 2.5 minutes). After all, I understand the first approach perfectly. But for me, the problem lies in switching gears for many problems where one has to know a few ways of solving the same thing.

This conversation is not about efficiency. Before I can go with and choose an efficient approach, I must know what other (or inefficient) approaches are. Logic of inefficient approach for one problem may be a highly efficient one for another problem.

Anyway, would be great to hear your view how we get to two different results - that is all I want to know. As it stands right now, either there is no second approach at all (in this case, seemingly we found a precedent to claim that the whole mathematical theory is not correct(!) ) or it is impossibly complicated .

Regards, _________________

There are times when I do not mind kudos...I do enjoy giving some for help

Re: Probability & Cominatorics Test #10 [#permalink]
14 Oct 2013, 09:30

Expert's post

obs23 wrote:

What I am concerned about at the end of the day, with all due respect to GMATTIGER and others agreeing with his solution, is that we have two different results. 10.91% and 12% (straight out, no rounding) - so I am puzzled, how is this possible (the way it stands right now, is that GMATTIGER is not correct, even though he used the method skillfully)?! GMATTIGER must be missing something, and if the answer choices contained 11% as an answer, he would simply be wrong. I just want to get to the same answer both ways and understand where he is incorrect (I just can't see what he is missing! [we are using probability theory that supposed to be correct!]) - and I mentioned above, I understand the second approach is more difficult one (the problem is I began to think in described above terms right away and by the time I realized I need turn other way, I have already spent 2.5 minutes). After all, I understand the first approach perfectly. But for me, the problem lies in switching gears for many problems where one has to know a few ways of solving the same thing.

This conversation is not about efficiency. Before I can go with and choose an efficient approach, I must know what other (or inefficient) approaches are. Logic of inefficient approach for one problem may be a highly efficient one for another problem.

Anyway, would be great to hear your view how we get to two different results - that is all I want to know. As it stands right now, either there is no second approach at all (in this case, seemingly we found a precedent to claim that the whole mathematical theory is not correct(!) ) or it is impossibly complicated .

Regards,

Dear obs23, Here's what I respected most in GMAT TIGER's response:

GMAT TIGER wrote:

I am little skeptical about the question. The question is silent about the order of outcome

. The root of the problem is that the question itself is poorly worded. This would never fly on the GMAT. That's precisely why folks are getting different denominators --- because the ambiguity of the question allows for multiple interpretations. A high quality GMAT math question involves no ambiguity. This is a low quality question.

Interpretation #1: First number can be any of 10 choices, and the second number can be any of 100 choices, for a 100 possibilities total (you could imagine them laid out in a 10 x 10 square). This method counts (23, 21) and (21, 23) as two different options. This gives the neat 12% answer.

Interpretation #2: This method does not count the same pair twice. It involves a "half square", counting only the diagonal and points below the diagonal. This one has 10C2 + 10 = 55 possible pairs, and this one leads to the 10.91% answer.

I prefer interpretation #2, but technically, there's nothing in the question that definitively allows us to determine whether the question means interpretation #1 or #2. Again, this is an essential failure on the part of the author of the question. In fact, whenever there's wide disagreement on the answer to a math question, chances are very very good it's because the question is not well-written.

Re: Probability & Cominatorics Test #10 [#permalink]
15 Oct 2013, 00:23

Expert's post

obs23 wrote:

Quote:

Let me ask you a question: does 10C2 give the following possible pairs: (20, 20), (21, 21), ..., (29, 29) ?

Apparently, it does not, but it should...That is a good question, why does not this formula include them? Or in other words, what does my 10C2 assume? I kind of see it is not correct and the subtle difference but I am afraid I am unable to articulate it to myself.

The way I see it now is that you have to think about this problem not in terms of one set of 10 numbers but somehow in terms of two different sets of identical numbers? Then it kind of makes sense with 10x10 / 2! - I still need a proper way to think about it, otherwise tomorrow I wont be able to solve a similar problem.

Either way, I also do not see clearly the problem with 10C2.

Looks like the good news is that I am correct on the second part: 2C1 * 3C1.

P.S. From what opera of combinatorics is this? I mean what is the concept behind this particular stuff? Kudos to the problem!

Responding to a pm:

The reason you don't get the same answer in the combination approach is that the combination approach (nCr) is incorrect here. The question specifically mentions that the two numbers picked can be the same. When we say nCr, we say that we pick r items out of n items. This means that once you pick an item, you cannot re-pick it. 10C2 = 10*9/2 Why? First item can be picked in 10 ways. Second item can be picked in 9 ways (assumed to be different from first). Since there is no order, we divide 10*9 by 2 to un-arrange it.

Since the items can be re-picked here, using nCr is incorrect. If you want to find the total number of combinations, you can use the basic counting principle. i.e. pick first number in 10 ways and second in 10 ways. Now divide 10*10 by 2 (because order does not matter - 21, 23 is the same as 23, 21) to get a total of 50 ways.

Required probability = 6/50 (correct answer)

In this way, probability and combinations approach give the same answer. _________________

Re: Probability & Cominatorics Test #10 [#permalink]
15 Oct 2013, 08:49

1

This post received KUDOS

Expert's post

VeritasPrepKarishma wrote:

Responding to a pm:

The reason you don't get the same answer in the combination approach is that the combination approach (nCr) is incorrect here. The question specifically mentions that the two numbers picked can be the same. When we say nCr, we say that we pick r items out of n items. This means that once you pick an item, you cannot re-pick it. 10C2 = 10*9/2 Why? First item can be picked in 10 ways. Second item can be picked in 9 ways (assumed to be different from first). Since there is no order, we divide 10*9 by 2 to un-arrange it.

Since the items can be re-picked here, using nCr is incorrect. If you want to find the total number of combinations, you can use the basic counting principle. i.e. pick first number in 10 ways and second in 10 ways. Now divide 10*10 by 2 (because order does not matter - 21, 23 is the same as 23, 21) to get a total of 50 ways.

Required probability = 6/50 (correct answer)

In this way, probability and combinations approach give the same answer.

Dear Karishma,

With all due respect, if we use the FCP, then 10*10 = 100, but it seems to me that not all of those will need to be divided by two. Of those 100 pairs, 90 of them are pairs with two different entries, and as you say, these are double-counted, so we need to divide by 2 ---- e.g., of course we don't want to count (21, 23) and (23, 21) twice. But, the ten pairs with two of the same number, e.g. (24, 24) ---- that particular pair does not get counted twice in the original 100, because there's only one way this could show up. If we divide all 100 by 2, I submit that this diagonal of doubles of the same number gets cheated, and only half of it counts, even though all of those double-pairs should count. I thought GMAT TIGER's elegant formulation of 10C2 + 10 parsed the situation efficiently. What do you think?

Re: Probability & Cominatorics Test #10 [#permalink]
15 Oct 2013, 11:15

Expert's post

mikemcgarry wrote:

VeritasPrepKarishma wrote:

Responding to a pm:

The reason you don't get the same answer in the combination approach is that the combination approach (nCr) is incorrect here. The question specifically mentions that the two numbers picked can be the same. When we say nCr, we say that we pick r items out of n items. This means that once you pick an item, you cannot re-pick it. 10C2 = 10*9/2 Why? First item can be picked in 10 ways. Second item can be picked in 9 ways (assumed to be different from first). Since there is no order, we divide 10*9 by 2 to un-arrange it.

Since the items can be re-picked here, using nCr is incorrect. If you want to find the total number of combinations, you can use the basic counting principle. i.e. pick first number in 10 ways and second in 10 ways. Now divide 10*10 by 2 (because order does not matter - 21, 23 is the same as 23, 21) to get a total of 50 ways.

Required probability = 6/50 (correct answer)

In this way, probability and combinations approach give the same answer.

Dear Karishma,

With all due respect, if we use the FCP, then 10*10 = 100, but it seems to me that not all of those will need to be divided by two. Of those 100 pairs, 90 of them are pairs with two different entries, and as you say, these are double-counted, so we need to divide by 2 ---- e.g., of course we don't want to count (21, 23) and (23, 21) twice. But, the ten pairs with two of the same number, e.g. (24, 24) ---- that particular pair does not get counted twice in the original 100, because there's only one way this could show up. If we divide all 100 by 2, I submit that this diagonal of doubles of the same number gets cheated, and only half of it counts, even though all of those double-pairs should count. I thought GMAT TIGER's elegant formulation of 10C2 + 10 parsed the situation efficiently. What do you think?

Respectfully, Mike

Yes Mike, let me revise the explanation and give some more details. Let me also add that this explanation is to the best of my understanding of the muddy waters of P&C.

As discussed before, the probability approach gives the answer as .12 which is correct. We say there is a sequence to picking the numbers 3/10*2/10 + 2/10*3/10 and we take care of it.

The combinations approach also gives the answer as .12 and is also correct. But we cannot un-arrange the selection, I agree. The combinations approach should be something like this: Favorable cases: 3C1*2C1 + 2C1*3C1 (you select the multiple of 3 first in 3 ways and then the prime number in 2 ways OR select the prime number first in 2 ways and then the multiple of 3 in 3 ways) Favorable cases: 12 (21, 23), (21, 29), (24, 23), (24, 29), (27, 23), (27, 29), (23, 21), (29, 21), (23, 24), (29, 24), (23, 27), (29, 27)

Total number of cases = 10*10 (Select the first number and then the second. The reason we cannot un-arrange it is that we can pick the same number again. So say talking in worldly terms, we picked 20, to pick 20 again, we will need to drop it and pick it again i.e. there will be a natural sequence to picking the numbers. We get 100 cases as (20, 20), (20, 21)....... (20, 29) (21, 20), (21, 21)........(21, 29) . . .

Each one of these cases is distinct and the probability of getting any one of these cases is the same. Hence required probability = 12/100 = 6/50

So probability and combinations will give the same answer.

Why is 6/45 incorrect? It doesn't include the cases where both numbers are the same and assumes that they have to be different. That is not correct. This answers obs23's question.

Why is 6/55 incorrect? The probability of achieving the 55 cases is not the same (It's similar to the throwing two dice situation if someone has come across those questions). The probability of achieving each of the favorable 6 un-arranged cases is the same i.e. you can get each of them in 2 ways. But the probability of achieving each of the total 55 un-arranged cases is not the same. You can get (20, 21) in 2 ways but (20, 20) in only one way in the un-arranged case. Hence you cannot add them up to get the total number of cases as 55. In total number of cases, the probability of achieving each case has to be the same. In case it isn't, we need to consider the order to get all the cases. And this is the reason I do not agree with GMAT TIGER's response.