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# Set S consists of all prime integers less than 10. If two

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Set S consists of all prime integers less than 10. If two [#permalink]

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15 Dec 2011, 18:56
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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

Last edited by Bunuel on 04 Feb 2012, 12:59, edited 1 time in total.
Edited the question
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Kudos [?]: 20 [0], given: 5

Re: probabilty [#permalink]

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15 Dec 2011, 19:22
S = {2,3,5,7}

chosen numbers =x,y
not chosen ones = a,b
such that xy < ab

chosen nos = a,b
not chosen ones = x,y
ab>xy

so the probability that product of chosen ones is more than not chosen ones is 1/2

[Reveal] Spoiler:
Option C
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Re: probabilty [#permalink]

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15 Dec 2011, 22:57
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The set will be
S = {2, 3, 5, 7}
Two numbers can be chosen from the set of four numbers in 4C2 = 6 ways.
Chosen Numbers ----- Product ----- Numbers not chosen ----- Product
2, 3 -------------------- 6 --------------- 5, 7 -------------------- 35
2, 5 -------------------- 10 --------------- 3, 7 -------------------- 21
2, 7 -------------------- 14 --------------- 3, 5 -------------------- 15
3, 5 -------------------- 15 --------------- 2, 7 -------------------- 14
3, 7 -------------------- 21 --------------- 2, 5 -------------------- 10
5, 7 -------------------- 35 --------------- 2, 3 -------------------- 6

Thus, from the table above, it is obvious that there are just three possible combinations in which the product of the numbers chosen is greater than the product of the numbers not chosen.
Thus the required probability = 3/6 = 1/2

Hope this is helpful.
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Kudos [?]: 13 [0], given: 18

Re: probabilty [#permalink]

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16 Dec 2011, 00:33
IMO C.

Went with the same approach but i stopped after 4th step since 5 and 6 will definitely be greater in the chosen values side as the values are increasing that side and decreasing in the other side
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Re: Set S consists of all prime integers less than 10. If two [#permalink]

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04 Feb 2012, 13:01
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ashiima 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) $$\frac{4}{5}$$

Shortcut solution:

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.

Hope it's clear.
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Re: Set S consists of all prime integers less than 10. If two [#permalink]

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24 May 2013, 04:42
Bumping for review and further discussion.
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Re: Set S consists of all prime integers less than 10. If two [#permalink]

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06 Mar 2015, 10:25
Hello from the GMAT Club BumpBot!

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Re: Set S consists of all prime integers less than 10. If two [#permalink]

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20 Jul 2016, 00:51
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: Set S consists of all prime integers less than 10. If two   [#permalink] 20 Jul 2016, 00:51
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# Set S consists of all prime integers less than 10. If two

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