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m10 #14 another solution

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m10 #14 another solution [#permalink] New post 15 Nov 2009, 02:55
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Found the OE bit complicated than usual probability q approach... so posting another solution which maybe easier to understand for some (who fall in similar category:) )

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}

Total ways to choose 2 numbers (repeat doesn't matter): 4x4=16
This would have been 4x3=12 had the Q not mentioned "not necessarily different". I missed that part in the test :(
Result is odd only when one number is 2 and the other any of the others.

Favorable cases when "2" is the first to be drawn {(2,3),(2,5),(2,7)}=3 , another 3 when "2" is the second to be drawn.

So, probability=(favorable cases/total cases)=6/16=3/8
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Re: m10 #14 another solution [#permalink] New post 15 Nov 2009, 03:14
Simple Explanation, Kudos !!
I hate Probability :evil:
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Re: m10 #14 another solution [#permalink] New post 15 Nov 2009, 06:14
ctrlaltdel wrote:
Simple Explanation, Kudos !!
I hate Probability :evil:


Thanks :-D
don't like it either, but something unavoidable!
Re: m10 #14 another solution   [#permalink] 15 Nov 2009, 06:14
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