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

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

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New post Updated on: 01 Mar 2018, 21:33
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C
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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) \(\frac{4}{5}\)

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

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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]

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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:

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: Prime integer probability  [#permalink]

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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 numbers less than 10. If two num  [#permalink]

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New post 07 Aug 2017, 18:13
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Re: Set S consists of all prime numbers less than 10. If two num   [#permalink] 07 Aug 2017, 18:13
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