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m15,#10

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28 Nov 2008, 01:10
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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) $$\frac{4}{5}$$

[Reveal] Spoiler: OA
C

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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28 Nov 2008, 02:21
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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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21 Dec 2009, 06:59
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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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30 Dec 2010, 01:12
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Expert's post
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) $$\frac{4}{5}$$

[Reveal] Spoiler: OA
C

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).

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03 Jan 2011, 12:17
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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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28 Nov 2008, 02:31
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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28 Nov 2008, 04:16
scthakur, nice explanation

C
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07 Dec 2008, 19:14
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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21 Dec 2009, 06:37
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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21 Dec 2009, 19:29
Ans : C

there are only two possibal out come and have 50% chance of geting either one
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28 Dec 2010, 09:56
C for me as well.
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28 Dec 2010, 20:36
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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28 Dec 2010, 23:19
GUD eXPLANATION..
its C.
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29 Dec 2010, 23:27
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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30 Dec 2011, 06:06
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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30 Dec 2011, 06:14
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
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30 Dec 2011, 07:30
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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30 Dec 2011, 18:05
C.
The Kaplan Instructor explained it pretty well.
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31 Dec 2011, 04:27
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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