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# M15-10

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Math Expert
Joined: 02 Sep 2009
Posts: 52401

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15 Sep 2014, 23:55
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Difficulty:

45% (medium)

Question Stats:

70% (01:14) correct 30% (02:02) wrong based on 86 sessions

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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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Math Expert
Joined: 02 Sep 2009
Posts: 52401

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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).

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Joined: 03 Jun 2013
Posts: 19
Concentration: Strategy, General Management
GMAT 1: 520 Q38 V32
GMAT 2: 530 Q44 V22
GMAT 3: 670 Q47 V34
WE: Information Technology (Consulting)

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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).

Hi,

Can u suggest another way to solve this problem.
Intern
Joined: 22 Oct 2014
Posts: 3
Location: India
Concentration: Entrepreneurship, Marketing
GMAT 1: 730 Q50 V40
WE: Sales (Energy and Utilities)

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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
Intern
Joined: 21 May 2013
Posts: 7

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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
Current Student
Joined: 29 Apr 2015
Posts: 26
Location: Russian Federation
GMAT 1: 710 Q48 V38
GPA: 4

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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.
Manager
Joined: 08 Jan 2013
Posts: 107

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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
Re: M15-10 &nbs [#permalink] 29 Oct 2018, 14:54
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# M15-10

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