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]

10 Aug 2009, 08:52

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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?

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

How come when I list them out and count them manually I don't get the correct answer?

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Re: Probability [#permalink]

10 Aug 2009, 11:49

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bipolarbear wrote:

As bkparikh mentioned I also get 3/8.
May be a hint, you might be counting (2,3) and (3,2) as different sets, but for such addition problems both represent the same set i.e. we have to count these as only one possibility.

Or may be you might be missing the fact that first number can be selected in four ways and the other can ALSO be selected in four ways (as it can be with repetition as per the question ). So 4 options for first one and 4 options for second one. Now since the order does not matter we get a total of 4*4/2! = 8 ways to select two numbers with repetition.

Hence our denominator ( total outcomes ) = 8.
For numerator ( favorable outcomes ) I did it manually and got 3 as (2,3),(2,5) and (2,7).
So, ans should be 3/8

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Re: Probability [#permalink]

08 Sep 2009, 14:05

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C

Put 16 in the denominator for 16 possibilities (4*4).

For the numerator you have 6 different ways you can get an odd sum:
2,3
2,5
2,7
3,2
5,2
7,2

So you have 6/16.

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Re: Probability [#permalink]

10 Sep 2009, 06:56

i am also gettin 3/8...what is the official answer.........

Re: Probability [#permalink] 10 Sep 2009, 06:56

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

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

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