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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] New post 30 Oct 2008, 03:42
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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?

* 1/8
* 1/6
* 3/8
* 1/2
* 5/8
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Re: PS : Probability of Odd sum [#permalink] New post 30 Oct 2008, 04:03
is the anaswer 1/2.

approach taken - prime num <10 (2,3,5,7)
prob odd = sets considered (2,3) , (2,5) and (2,7)

prob (2,3) = 1/4 * 1/3 * 2 (order may be repeated). Thus for three times = 1/12 *2*3 = 1/2

Wats the OA?
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Re: PS : Probability of Odd sum [#permalink] New post 30 Oct 2008, 06:54
imo - 1/2


total possibilities - 4c2= 6 ..( 2 prime no out of 2,3,5,and 7 )
prob odd = 3 ways ( sets considered (2,3) , (2,5) and (2,7) )

3/6=1/2 .
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Re: PS : Probability of Odd sum [#permalink] New post 30 Oct 2008, 08:14
3/8

As i understand(from "not necessarily different"), the selected number is put into the set before second pick.

so probability reduces to 1/4 *3/4 *2 = 3/8
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Re: PS : Probability of Odd sum [#permalink] New post 30 Oct 2008, 08:35
mbajingle wrote:
3/8

As i understand(from "not necessarily different"), the selected number is put into the set before second pick.

so probability reduces to 1/4 *3/4 *2 = 3/8


1/4 *3/4 *2 = 3/8 ------------- how did you arrive at this ??
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Re: PS : Probability of Odd sum [#permalink] New post 30 Oct 2008, 08:49
amitdgr wrote:
mbajingle wrote:
3/8

As i understand(from "not necessarily different"), the selected number is put into the set before second pick.

so probability reduces to 1/4 *3/4 *2 = 3/8


1/4 *3/4 *2 = 3/8 ------------- how did you arrive at this ??

A 2 and an odd prime is must if sum of two prime number needs to be zero. So probability of picking 2 is 1/4. If you put back 2 in the set before selecting second number, the probability of selecting odd number during second pick is 3/4. 2 can be picked either in first or second pick. Thus probability of favorable event is 2*1/4*3/4.

What is OA?
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Re: PS : Probability of Odd sum [#permalink] New post 30 Oct 2008, 08:55
mbajingle wrote:
A 2 and an odd prime is must if sum of two prime number needs to be zero. So probability of picking 2 is 1/4. If you put back 2 in the set before selecting second number, the probability of selecting odd number during second pick is 3/4. 2 can be picked either in first or second pick. Thus probability of favorable event is 2*1/4*3/4.

What is OA?


OA is 3/8 ....

I thought 2+3 and 3+2 is same ... apparently order matters here .... how do we know that the order matters here ??
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Re: PS : Probability of Odd sum [#permalink] New post 30 Oct 2008, 09:17
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amitdgr wrote:
mbajingle wrote:
A 2 and an odd prime is must if sum of two prime number needs to be zero. So probability of picking 2 is 1/4. If you put back 2 in the set before selecting second number, the probability of selecting odd number during second pick is 3/4. 2 can be picked either in first or second pick. Thus probability of favorable event is 2*1/4*3/4.

What is OA?


OA is 3/8 ....

I thought 2+3 and 3+2 is same ... apparently order matters here .... how do we know that the order matters here ??


The first and the second pick are two different unrelated(outcome of second doesn't depend on first) events. So 3+2 is different from 2+3.
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Re: PS : Probability of Odd sum [#permalink] New post 30 Oct 2008, 09:24
mbajingle wrote:
The first and the second pick are two different unrelated(outcome of second doesn't depend on first) events. So 3+2 is different from 2+3.


Piffany !! Thanks mbajingle :) +1 for you
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Re: PS : Probability of Odd sum [#permalink] New post 30 Oct 2008, 09:55
another method is:
total outcomes=4^2
favourable outcomes=6
p(odd sum)=6/4^2=3/8
Re: PS : Probability of Odd sum   [#permalink] 30 Oct 2008, 09:55
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