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What is the probability that rs=r?

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What is the probability that rs=r? [#permalink]

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Sets R and S each contain three distinct positive integers. If integer r is randomly selected from R and integer s is randomly selected from S, what is the probability that rs = r?

(1) The probability that rs = s is 1/3

(2) The probability that r + s = 2 is 1/9
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Re: What is the probability that rs=r? [#permalink]

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New post 12 Jun 2013, 14:46
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Sets R and S each contain three distinct positive integers. If integer r is randomly selected from R and integer s is randomly selected from S, what is the probability that rs = r?

\(rs = r\) --> \(r(s-1)=0\). This will be true if \(r=0\) or \(s=1\). We are told that the sets contain positive integers, thus this will be true only if \(s=1\). So, the question basically asks what is the probability that integer selected from S is 1.

(1) The probability that rs = s is 1/3. \(rs = s\) --> \(s(r-1)=0\). The the probability that 1 is selected from R 1/3. Not sufficient.

(2) The probability that r + s = 2 is 1/9. Since the sets contain positive integers, then r=s=1. So, we have that the probability that 1 is selected from R and 1 is selected from S is 1/9 --> P(1 and 1)=P(1 from R)*P(1 from S)=1/3*1/3=1/9 (since each set contains three distinct integers). Therefore the probability that integer selected from S is 1 is 1/3. Sufficient.

Answer: B.

Hope it's clear.
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Re: What is the probability that rs=r? [#permalink]

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New post 12 Jun 2013, 14:50
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Bunuel wrote:
Sets R and S each contain three distinct positive integers. If integer r is randomly selected from R and integer s is randomly selected from S, what is the probability that rs = r?

\(rs = r\) --> \(r(s-1)=0\). This will be true if \(r=0\) or \(s=1\). We are told that the sets contain positive integers, thus this will be true only if \(s=1\). So, the question basically asks what is the probability that integer selected from S is 1.

(1) The probability that rs = s is 1/3. \(rs = s\) --> \(s(r-1)=0\). The the probability that 1 is selected from R 1/3. Not sufficient.

(2) The probability that r + s = 2 is 1/9. Since the sets contain positive integers, then r=s=1. So, we have that the probability that 1 is selected from R and 1 is selected from S is 1/9 --> P(1 and 1)=P(1 from R)*P(1 from S)=1/3*1/3=1/9 (since each set contains three distinct integers). Therefore the probability that integer selected from S is 1 is 1/3. Sufficient.

Answer: B.

Hope it's clear.


Thank you. I enjoy reading your explanations. They are very clear.
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Re: What is the probability that rs=r? [#permalink]

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New post 09 Jul 2013, 12:14
1
Bunuel wrote:
Sets R and S each contain three distinct positive integers. If integer r is randomly selected from R and integer s is randomly selected from S, what is the probability that rs = r?

\(rs = r\) --> \(r(s-1)=0\). This will be true if \(r=0\) or \(s=1\). We are told that the sets contain positive integers, thus this will be true only if \(s=1\). So, the question basically asks what is the probability that integer selected from S is 1.

(1) The probability that rs = s is 1/3. \(rs = s\) --> \(s(r-1)=0\). The the probability that 1 is selected from R 1/3. Not sufficient.

(2) The probability that r + s = 2 is 1/9. Since the sets contain positive integers, then r=s=1. So, we have that the probability that 1 is selected from R and 1 is selected from S is 1/9 --> P(1 and 1)=P(1 from R)*P(1 from S)=1/3*1/3=1/9 (since each set contains three distinct integers). Therefore the probability that integer selected from S is 1 is 1/3. Sufficient.

Answer: B.

Hope it's clear.



Hi Bunnel ,

Can you please clarify one thing ,we are breaking the probability (1/9) in the form 1/3 and 1/3 becoz we have 3 elements in the set :-

Or else if the no of elements wasn't given then we could have broken the same in various other ways like 1/6 *2/3 or other ,then b would also have been insufficient.

Is my understanding correct ?

Thanks in advance.

Countdown
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Re: What is the probability that rs=r? [#permalink]

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New post 09 Jul 2013, 23:57
2
Countdown wrote:
Bunuel wrote:
Sets R and S each contain three distinct positive integers. If integer r is randomly selected from R and integer s is randomly selected from S, what is the probability that rs = r?

\(rs = r\) --> \(r(s-1)=0\). This will be true if \(r=0\) or \(s=1\). We are told that the sets contain positive integers, thus this will be true only if \(s=1\). So, the question basically asks what is the probability that integer selected from S is 1.

(1) The probability that rs = s is 1/3. \(rs = s\) --> \(s(r-1)=0\). The the probability that 1 is selected from R 1/3. Not sufficient.

(2) The probability that r + s = 2 is 1/9. Since the sets contain positive integers, then r=s=1. So, we have that the probability that 1 is selected from R and 1 is selected from S is 1/9 --> P(1 and 1)=P(1 from R)*P(1 from S)=1/3*1/3=1/9 (since each set contains three distinct integers). Therefore the probability that integer selected from S is 1 is 1/3. Sufficient.

Answer: B.

Hope it's clear.



Hi Bunnel ,

Can you please clarify one thing ,we are breaking the probability (1/9) in the form 1/3 and 1/3 becoz we have 3 elements in the set :-

Or else if the no of elements wasn't given then we could have broken the same in various other ways like 1/6 *2/3 or other ,then b would also have been insufficient.

Is my understanding correct ?

Thanks in advance.

Countdown


Notice that not only we are told that the sets contain 3 integers each but we are also told that each set contains three distinct positive integers. From (2) we have that R={1, a, b} and S={1, c, d}, where neither from a, b, c, and d is 1. So, P(1 from R)=P(1 from S)=1/3.

Hope it's clear.
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Re: What is the probability that rs=r? [#permalink]

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New post 15 Aug 2013, 22:10
1
Bunuel wrote:
Sets R and S each contain three distinct positive integers. If integer r is randomly selected from R and integer s is randomly selected from S, what is the probability that rs = r?


I know i am wrong but kindly do explain why cant i read the statement above as R and S both contain Three distinct positive integers.

R = {a, b , c}
S = {p, q, r}

Is it because its not explicitly mentioned that there is no overlap?
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Re: What is the probability that rs=r? [#permalink]

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New post 16 Aug 2013, 01:47
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Transcendentalist wrote:
Bunuel wrote:
Sets R and S each contain three distinct positive integers. If integer r is randomly selected from R and integer s is randomly selected from S, what is the probability that rs = r?


I know i am wrong but kindly do explain why cant i read the statement above as R and S both contain Three distinct positive integers.

R = {a, b , c}
S = {p, q, r}

Is it because its not explicitly mentioned that there is no overlap?


Sorry, but I don't understand your questions at all...
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Re: What is the probability that rs=r? [#permalink]

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New post 16 Aug 2013, 02:30
1
Bunuel wrote:
Transcendentalist wrote:
Bunuel wrote:
Sets R and S each contain three distinct positive integers. If integer r is randomly selected from R and integer s is randomly selected from S, what is the probability that rs = r?


I know i am wrong but kindly do explain why cant i read the statement above as R and S both contain Three distinct positive integers.

R = {a, b , c}
S = {p, q, r}

Is it because its not explicitly mentioned that there is no overlap?


Sorry, but I don't understand your questions at all...


Sorry i had a mistake in my last post...Why cant i Assume that

R = {r, a, b}
S = {s, c, d}

Where a,b,c,d,r,s are all distinct?

If that is the case then

We have to check if s= 1

A - s(r-1) = 0 s cant be zero so r is 1.. If r is 1, s cant be one as all are distinct - Sufficient

B - r + s = 2 1/9 Impossible hence insufficient?
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Economist GMAT Test - 730, Q50, V41 Aug 9th, 2013
Manhattan GMAT Test - 670, Q45, V36 Aug 11th, 2013
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GmatPrep CAT 1 - 770, Q50, V44 Aug 24th, 2013
Manhattan GMAT Test - 690, Q45, V39 Aug 30th, 2013
Manhattan GMAT Test - 710, Q48, V39 Sep 13th, 2013
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GMAT - 770, Q50, V44, Oct 7th, 2013
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Re: What is the probability that rs=r? [#permalink]

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New post 16 Aug 2013, 02:36
Transcendentalist wrote:
Bunuel wrote:
Transcendentalist wrote:
Sets R and S each contain three distinct positive integers. If integer r is randomly selected from R and integer s is randomly selected from S, what is the probability that rs = r?


I know i am wrong but kindly do explain why cant i read the statement above as R and S both contain Three distinct positive integers.

R = {a, b , c}
S = {p, q, r}

Is it because its not explicitly mentioned that there is no overlap?


Sorry, but I don't understand your questions at all...


Sorry i had a mistake in my last post...Why cant i Assume that

R = {r, a, b}
S = {s, c, d}

Where a,b,c,d,r,s are all distinct?

If that is the case then

We have to check if s= 1

A - s(r-1) = 0 s cant be zero so r is 1.. If r is 1, s cant be one as all are distinct - Sufficient

B - r + s = 2 1/9 Impossible hence insufficient?


We are told that sets R and S each contain three distinct positive integers. This does not mean that all 6 integers are distinct.
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Re: What is the probability that rs=r? [#permalink]

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New post 11 Oct 2015, 09:41
We are told that sets R and S each contain three distinct positive integers. This does not mean that all 6 integers are distinct.:

So why is 1/infinity not the accepted sol. for st. 1 so ans. be D?
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Re: What is the probability that rs=r? [#permalink]

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New post 11 Oct 2015, 09:43
Vale1Spa wrote:
We are told that sets R and S each contain three distinct positive integers. This does not mean that all 6 integers are distinct.:

So why is 1/infinity not the accepted sol. for st. 1 so ans. be D?


Your question is not clear. Also, note that infinity is not a number.
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Re: What is the probability that rs=r? [#permalink]

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New post 11 Oct 2015, 10:03
1 is one of the infinite numbers from which S is formed.
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Re: What is the probability that rs=r? [#permalink]

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Re: What is the probability that rs=r? [#permalink]

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New post 11 Oct 2015, 10:48
Anyway, limit 1/infin. =0 is never accepted as answer by GMAT. The datae were not clear, imho.
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Re: What is the probability that rs=r? [#permalink]

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New post 12 Oct 2015, 11:17
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Sets R and S each contain three distinct positive integers. If integer r is randomly selected from R and integer s is randomly selected from S, what is the probability that rs = r?

(1) The probability that rs = s is 1/3

(2) The probability that r + s = 2 is 1/9

We can modify the question into: rs=r, rs-r=0. r(s-1)=0,and ultimately whether r=0 or s=1? But the question is regarding distinct positive integers, so what we really want to know is whether s=1.
For condition 2, the only situation where r+s=2 is when r=s=1, so the probability that s=1 is only 1/9, making the condition unique and sufficient,
Whereas condition 1 gives s(r-1)=0 and the probability that r=1 is 1/3, but we cannot know the probability that s=1, so this condition is insufficient, making the answer (B).

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Re: What is the probability that rs=r? [#permalink]

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Re: What is the probability that rs=r?   [#permalink] 08 Feb 2018, 14:25
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