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.
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.
