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M15-10

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M15-10  [#permalink]

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New post 15 Sep 2014, 23:55
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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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Re M15-10  [#permalink]

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New post 15 Sep 2014, 23:55
1
Official Solution:

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. \(\frac{4}{5}\)


\(S=\{2, 3, 5, 7\}\)

The simplest way to solve this question 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 \(\frac{1}{2}\) (the probability doesn't favor any of two subgroups).


Answer: C
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Collection of Questions:
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Re: M15-10  [#permalink]

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New post 16 Dec 2014, 22:47
Bunuel wrote:
Official Solution:

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. \(\frac{4}{5}\)


\(S=\{2, 3, 5, 7\}\)

The simplest way to solve this question 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 \(\frac{1}{2}\) (the probability doesn't favor any of two subgroups).


Answer: C


Hi,

Can u suggest another way to solve this problem.
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Re: M15-10  [#permalink]

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New post 25 Dec 2014, 10:52
3
Prime numbers less than 10: 2,3,5,7==>4 numbers in set S

2 are selceted==> 4!/2!*2!= 6 ways of selecting two numbers from set S

Favourable outcomes: 3-5,3-7,5-7==>3

P=3/6 =1/2 Ans. C
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Re: M15-10  [#permalink]

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New post 07 Jan 2015, 23:13
Set : 2,3,5,7.

Recognise that all combinations involving 3,5,7 form a product greater then the other numbers not chosen.

Probability of picking 3,5 or 7 as the first number : 3/4
Probability of picking 3,5 or 7 as the second number : (3-1)/(4-1) = 2/3

3/4 * 2/3 = 1/2
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M15-10  [#permalink]

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New post 12 Mar 2016, 07:13
1
Bunuel 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. \(\frac{4}{5}\)


To solve this problem it's important to understand that there cannot be a situation in which product of two chosen integers is equal to product of two left integers. It's can be concluded from the fact that all integers in the set are prime integers.

There can be 6 different products, since 4C2 = 6. And in 3 cases product of chosen integers will be greater than product of left integers and in 3 cases product of chosen integers will be less than product of left integers.

Therefore the probability is exactly 1/2.
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Re: M15-10  [#permalink]

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New post 29 Oct 2018, 14:54
Total outcome i.e. selecting 2 numbers out of 4 -> 4C2

For product of selected numbers to be greater, neither number should be 2 as max product for 2 = 2*7 => 14 which is less than min product for next greater number 3 i.e. 3*5 => 15.
So both numbers should be 3,5 or 7.

So, probability for selecting 2 numbers with greater product = 3C2 / 4C2 i.e. selecting 2 numbers from 3,5,7 divided by Total possible selections => 1/2
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Re: M15-10 &nbs [#permalink] 29 Oct 2018, 14:54
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