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# m15 10

Author Message
Director
Joined: 01 Apr 2008
Posts: 872

Kudos [?]: 860 [0], given: 18

Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014

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19 Sep 2009, 04:04
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?

I think it should be specified that two 'different' numbers are chosen. if not then 7,7 is also a possibility and then the probability will not be 1/2.

Kudos [?]: 860 [0], given: 18

Manager
Joined: 10 Jul 2009
Posts: 164

Kudos [?]: 120 [0], given: 8

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19 Sep 2009, 12:54
S = { 2,3,5,7}
Number of ways for chosing two numbers = 4c2 = 6
Number of out comes where product of two numbers is greater than product of remaining numbers = 3 (3*7, 5*7, 3*5)

So probability = 3/6 = 1/2

Different numbers need not be specified as the set has each number only one. If it is repeated many times then "different number" wording should be used.

Kudos [?]: 120 [0], given: 8

Intern
Joined: 23 Sep 2009
Posts: 12

Kudos [?]: [0], given: 0

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24 Sep 2009, 01:15
Set of prime numbers less than 10 is S = {2, 3, 5, 7}

Using counting methods, the number of ways to choose 2 items out of 4 is 4!/2!*2! = 6

The 6 possible products are:
2 x 3 = 6
2 x 5 = 10
2 x 7 = 14
3 x 5 = 15
3 x 7 = 21
5 x 7 = 35

... of which the last 3 products are larger than the first 3 products.

Therefore, the probability of selecting 2 numbers with a product that is larger than those not selected is 3/6 = 0.5

Kudos [?]: [0], given: 0

Re: m15 10   [#permalink] 24 Sep 2009, 01:15
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# m15 10

Moderator: Bunuel

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