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# Set S consists of all prime integers less than 10. If a

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Set S consists of all prime integers less than 10. If a [#permalink]  04 Sep 2008, 01:35
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Set S consists of all prime integers less than 10. If a number is selected from set S at random and then another number, not necessarily different, is selected from set S at random, what is the probability that the sum of these numbers is odd?

* $$\frac{1}{8}$$
* $$\frac{1}{6}$$
* $$\frac{3}{8}$$
* $$\frac{1}{2}$$
* $$\frac{5}{8}$$
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Schools: Wharton, LBS, UChicago, Kellogg MMM (Donald Jacobs Scholarship), Stanford, HBS
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Re: probability [#permalink]  04 Sep 2008, 01:57
3/8

Set S: 2, 3, 5, 7

Total cases: 4^2 = 16

First number 2: 1 case even (2,2), 3 cases odd (2,3 ; 2,5; 2,7)
First number odd (3 or 5 or 7): 3 cases even, 1 case odd

To sum it up:

EVEN: 1 + 3*3 = 10 out of 16
ODD: 3 + 1*3 = 6 out of 16

Pr(odd): 6/16 = 3/8
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Re: probability [#permalink]  04 Sep 2008, 06:38
arjtryarjtry wrote:
Set S consists of all prime integers less than 10. If a number is selected from set S at random and then another number, not necessarily different, is selected from set S at random, what is the probability that the sum of these numbers is odd?

* $$\frac{1}{8}$$
* $$\frac{1}{6}$$
* $$\frac{3}{8}$$
* $$\frac{1}{2}$$
* $$\frac{5}{8}$$

2,3,5,7 -- First selection set
2,3,5,7-- second selection set

to get the odd.. sum odd+even.

=(select 2 from 1st set * select any odd number from 2nd set) +(select 2 from 2st set * select any odd number from 1st set)
= 1*3+1*3 =6

p= 6/4*4 = 3/8
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Last edited by x2suresh on 04 Sep 2008, 07:05, edited 1 time in total.
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Re: probability [#permalink]  04 Sep 2008, 06:54
x2suresh wrote:
arjtryarjtry wrote:
Set S consists of all prime integers less than 10. If a number is selected from set S at random and then another number, not necessarily different, is selected from set S at random, what is the probability that the sum of these numbers is odd?

* $$\frac{1}{8}$$
* $$\frac{1}{6}$$
* $$\frac{3}{8}$$
* $$\frac{1}{2}$$
* $$\frac{5}{8}$$

2,3,5,7 -- First selection set
2,3,5,7-- second selection set

to get the odd.. sum odd+even.

=(select 2 from 1st set * select any odd number from 2nd set) +(select 2 from 1st set * select any odd number from 2nd set)
= 1*3+1*3 =6

p= 6/4*4 = 3/8

I believe you wanted to write
(select 2 from 1st set * select any odd number from 2nd set) +(select 2 from 2nd set * select any odd number from 1st set
Re: probability   [#permalink] 04 Sep 2008, 06:54
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