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# What is the probability that rs=r?

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Intern
Joined: 06 Jun 2013
Posts: 7
What is the probability that rs=r? [#permalink]

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12 Jun 2013, 14:29
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95% (hard)

Question Stats:

40% (01:30) correct 60% (01:45) wrong based on 536 sessions

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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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Joined: 02 Sep 2009
Posts: 46297
Re: What is the probability that rs=r? [#permalink]

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12 Jun 2013, 14:46
10
4
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.

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

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12 Jun 2013, 14:50
1
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.

Hope it's clear.

Intern
Joined: 16 Jan 2013
Posts: 30
Concentration: Finance, Entrepreneurship
GMAT Date: 08-25-2013
Re: What is the probability that rs=r? [#permalink]

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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.

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 ?

Countdown
Math Expert
Joined: 02 Sep 2009
Posts: 46297
Re: What is the probability that rs=r? [#permalink]

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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.

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 ?

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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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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Joined: 02 Sep 2009
Posts: 46297
Re: What is the probability that rs=r? [#permalink]

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16 Aug 2013, 01:47
1
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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Joined: 24 Nov 2012
Posts: 171
Concentration: Sustainability, Entrepreneurship
GMAT 1: 770 Q50 V44
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Re: What is the probability that rs=r? [#permalink]

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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?
_________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

http://www.manhattangmat.com/blog/index.php/author/cbermanmanhattanprep-com/ - This is worth its weight in gold

Economist GMAT Test - 730, Q50, V41 Aug 9th, 2013
Manhattan GMAT Test - 670, Q45, V36 Aug 11th, 2013
Manhattan GMAT Test - 680, Q47, V36 Aug 17th, 2013
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
GmatPrep CAT 2 - 740, Q49, V41 Oct 6th, 2013

GMAT - 770, Q50, V44, Oct 7th, 2013
My Debrief - http://gmatclub.com/forum/from-the-ashes-thou-shall-rise-770-q-50-v-44-awa-5-ir-162299.html#p1284542

Math Expert
Joined: 02 Sep 2009
Posts: 46297
Re: What is the probability that rs=r? [#permalink]

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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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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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Posts: 46297
Re: What is the probability that rs=r? [#permalink]

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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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11 Oct 2015, 10:03
1 is one of the infinite numbers from which S is formed.
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Posts: 46297
Re: What is the probability that rs=r? [#permalink]

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11 Oct 2015, 10:08
Vale1Spa wrote:
1 is one of the infinite numbers from which S is formed.

Sorry, I do not understand what you are trying to say...
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Re: What is the probability that rs=r? [#permalink]

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