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VP
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Set S consists of all prime integers less than 10. If two numbers are chosen form set S at random, what is the probability that the product of these numbers will be greater than the product of the numbers which were not chosen? (A) \frac{1}{3}(B) \frac{2}{3}(C) \frac{1}{2}(D) \frac{7}{10}(E) \frac{4}{5} Source: GMAT Club Tests - hardest GMAT questions What is wrong with my approach? The set consists of (2,3,5,7) Niow acc to soncition the nos satisfying are (5,7), (3,5) and (3,7) Thus the probabiliy is (1/4*1/4)+(1/4*1/4)+(1/4*1/4)=3/16 But this is not the ans 
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SVP
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It should be (1/4*1/3)+(1/4*1/3)+(1/4*1/3) = 3/12 = 1/4.
But, the sequence is immaterial here as we are interested in sum. Hence, (5,3) and (3,5) will give the same result. Thus, multiply 1/4 by 2 and 1/2 should be the answer.
Alternatively, I will solve it as follows:
favorable outcome = 3 (as you have listed earlier).
Since, order does not matter, total outcome = 4C2 = 6
Hence, probability = 3/6 = 1/2
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Manager
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I would go with exactly the same approach as the second one stated by scthakur.
Total outcomes = 2 numbers out of 4 hence 4C2 =6
Possible outcomes = 2 numbers out of 3 hence 3C2 = 3
Probability = 3/6 = 1/2
Whats the QA ?
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Manager
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scthakur, nice explanation
C
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Intern
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We now the total possibilities is 4c2
List out the possibilities
Chossen Not choosen
23 57 - no
25 37 -no
27 35 - no
35 27 -yes
37 25 - yes
57 23 - yes
4c2=6
3/6 = 1/2
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Manager
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The total number of combinations is the following: 4!/(2!*2!) = 6
The only ones that work are these: 3*5, 3*7, 5*7
3/6 = 1/2
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All solutions that do not include a 2 satisfy the condition. So basically we just need to figure out the odds of not drawing a 2. 3/4 * 2/3 = 6/12 = 1/2 So C.
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Manager
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Ans : C there are only two possibal out come and have 50% chance of geting either one
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C for me as well. 
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Intern
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i think the ans is 1/3
Pls check whether my understanding is correct
We need to select two numbers from 1, 2, 3 ,5, 7 ( left out are 4, 6, 8, 9)
There are only two possibility, either its (5,7) or (7,5), for which product is 35 and is greater than the product of (4,6) / (6,4) rest all are lower than any combination from left out series.
So p= 2/4C2 = 2/6 = 1/3 Ans
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Manager
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GUD eXPLANATION.. its C.
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C. The limitations of the set allow you to list the specific product options and compare them. I'm someone who's skilling up in quant and since I know that I make simple mistakes, I opted to actually write out the product options.
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ritula wrote: Set S consists of all prime integers less than 10. If two numbers are chosen form set S at random, what is the probability that the product of these numbers will be greater than the product of the numbers which were not chosen? (A) \frac{1}{3}(B) \frac{2}{3}(C) \frac{1}{2}(D) \frac{7}{10}(E) \frac{4}{5} Source: GMAT Club Tests - hardest GMAT questions S={2, 3, 5, 7} The simplest way would be to realize that we choose half of the numbers ( basically we divide the group of 4 into two smaller groups of 2) and since a tie is not possible then the probability that the product of the numbers in either of subgroup is more than that of in another is 1/2 (the probability doesn't favor any of two subgroups). Answer: C.
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Kaplan GMAT Instructor
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zahuruddin wrote: i think the ans is 1/3
Pls check whether my understanding is correct
We need to select two numbers from 1, 2, 3 ,5, 7 ( left out are 4, 6, 8, 9)
There are only two possibility, either its (5,7) or (7,5), for which product is 35 and is greater than the product of (4,6) / (6,4) rest all are lower than any combination from left out series.
So p= 2/4C2 = 2/6 = 1/3 Ans Hi! 1 isn't a prime number, which is the first reason that you're deriving a different answer. The second reason is that you've misinterpreted the question - we're not comparing the product of the two numbers we select to the product of the non-primes less than 10; we're comparing the product of the two numbers we select to the product of the other primes less than 10. So, our set is {2, 3, 5, 7} and we're comparing the products of two pairs of numbers from within the set. The quickest ways to solve are to either use Bunuel's intuitive approach (we're comparing 2 vs 2 with no ties, so symmetry will lead to half the pairs being greater than the other half) or via brute force. Since there are only 6 possibilities, it doesn't take long to count them out: 2*3 vs 5*7 - not greater 2*5 vs 3*7 - not greater 2*7 vs 3*5 - not greater 3*5 vs 2*7 - greater 3*7 vs 2*5 - greater 5*7 vs 3*5 - greater Probability = (# of desired outcomes)/(total # of possibilities) = 3/6 = 1/2
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Intern
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Counting 1 as a prime number is exactly the mistake I made... Funnily enough, they have an answer choice that corresponds to this particular way of "solving" the question.
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C
total ways of choosing 2 out of 4 = !4/(!2*!2) i.e. 6
total ways the product can be greater than the other two = (3,5), (5,7), (7,3) = 3
answer=3/6 = 1/2
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Manager
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I solved by calculating how many solutions do not involve 2.
1 - 1/4 * 1 - 1/3 = 6 /12 = 1/2. answer is C
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Manager
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C. The Kaplan Instructor explained it pretty well.
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Intern
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Moss wrote: All solutions that do not include a 2 satisfy the condition. So basically we just need to figure out the odds of not drawing a 2.
3/4 * 2/3 = 6/12 = 1/2
So C. This one is fast and least error prone for this problem. Thanks.
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