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Math Expert V
Joined: 02 Sep 2009
Posts: 58381

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Difficulty:   35% (medium)

Question Stats: 72% (01:18) correct 28% (02:02) wrong based on 93 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 V
Joined: 02 Sep 2009
Posts: 58381

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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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Intern  B
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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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  B
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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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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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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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  S
Joined: 08 Jan 2013
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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   [#permalink] 29 Oct 2018, 15:54
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