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

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ashiima
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Set S consists of all prime integers less than 10. If two numbers are [#permalink] New post   Updated on: 01 Mar 2018, 21:33
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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) \(\frac{4}{5}\)

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C

Originally posted by ashiima on 15 Dec 2011, 18:56.
Last edited by Bunuel on 01 Mar 2018, 21:33, edited 2 times in total.
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siddharthmuzumdar
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Re: Set S consists of all prime integers less than 10. If two numbers are [#permalink] New post   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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Re: Set S consists of all prime integers less than 10. If two numbers are [#permalink] New post   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:

Set \(S\) is defined as \(\{2, 3, 5, 7\}\).

The most straightforward way to address this question is to recognize that we are selecting half of the numbers from the set (essentially dividing the group of four into two smaller groups of two). As a tie is not possible, the probability that the product of the numbers in one subgroup exceeds that in the other subgroup is \(\frac{1}{2}\), because neither subgroup is favored over the other in terms of probability.

Answer: C.

Hope it's clear.
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Re: Set S consists of all prime integers less than 10. If two numbers are [#permalink] New post   07 Aug 2017, 18:13
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Primes in action -> 2,3,5,7

Cases ->
2,3 => (NO)
2,5 => (NO)
2,7 => (NO)
3,5 => (YES)
3,7 => (YES)
5,7 => (YES)


Total cases -> 6
Favourable cases -> 3

P(E) => 3/6 => 1/2

Hence C.
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Re: Set S consists of all prime integers less than 10. If two numbers are [#permalink] New post   11 Jan 2013, 13:00
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Prime integers below 10: 2, 3, 5, 7

Total outcomes: Number of ways you can choose 2 from 4 = \(4C2= 6 pairs\)
No. of ways for Event: Pairs that will have products greater than product of pairs not selected i.e. (3,5), (3,7), (5,7) \(= 3 pairs\)

Probability = number of ways/Total outcomes \(= 3/6 = 1/2\)

Hence choice(C) is correct.
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Re: Set S consists of all prime integers less than 10. If two numbers are [#permalink] New post   01 Feb 2020, 07:43
Considering the greatest (5,7) and the least (2,3) from total (2,3,5,7)

No numbers in combination both can be greater than product of the greatest ones i.e. 5*7 = 35

And thus rest 2 will remain least.

Thus, P = 2/4 = 1/2
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Re: Set S consists of all prime integers less than 10. If two numbers are [#permalink] New post   10 Jun 2023, 23:00
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Re: Set S consists of all prime integers less than 10. If two numbers are [#permalink]
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