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M10 Q#14

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M10 Q#14 [#permalink] New post 29 Aug 2010, 08:55
Set consists of all prime integers less than 10. If a number is selected from set at random and then another number, not necessarily different, is selected from set 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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For Ans: pls refer - probability-69758.html



The OC says - the total probability of getting an ODD sum =
Prob( picking up first number as 2 & 2nd number as ODD) + Prob( picking up first number as ODD & 2nd number as 2 )

I dont quite understand the logic here. For the actual operation - SUM , the order of picking up the numbers doesnot matter.
Should the ans then be 3/16 ?!

Please help
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Re: M10 Q#14 [#permalink] New post 29 Aug 2010, 09:22
The numbers in the set are : 2,3,5,7
The order matters, because 2 is the only even prime number.
Suppose, you have drawn 2 in the first attempt, now you have to get any odd number to make sure the sum is odd.
Similarly, if you draw any odd number in the first draw, you have to draw 2 to make the sum odd.

Hope that clears your doubt.
Thanks

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Re: M10 Q#14 [#permalink] New post 29 Aug 2010, 09:34
Sorry, I Still don't get it.
Since for the sum to be Odd, one number has to be even, in this case it is 2.
How can the order still matter in this case?

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Re: M10 Q#14 [#permalink] New post 29 Aug 2010, 17:32
wininblue wrote:
Set consists of all prime integers less than 10. If a number is selected from set at random and then another number, not necessarily different, is selected from set at random, what is the probability that the sum of these numbers is odd?

 1/8
 1/6
 3/8
 1/2
 5/8

Please help


set is 2,3,5,7

P(sum odd)= P(pick 2) and P(pick Other than 2) OR P(pick other than 2) and P(pick 2)

P(SumOdd) = \frac{1}{4}*\frac{3}{4} + \frac{3}{4}*\frac{1}{4}

therefore 3/16 + 3/16 = 6/16 = 3/8

hope it helps
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Re: M10 Q#14 [#permalink] New post 29 Aug 2010, 19:36
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As you pick the first number, it doesn't really matter which one you pick. The only thing the first number changes is what options you have when you pick the second number.

Two Possible Situations: 1st: Pick 2 and then forst 2nd you'd have to pick one of the odd numbers in order to get EVEN + ODD = ODD.

So, Chance of picking 2 on the first try? 1 out of 4 = \frac{1}{4}

Now on the second pick, we must have one of the odd numbers 3, 5, or 7, so 3 out of 4, or \frac{3}{4}. In order to figure out the chance of getting an odd when we pick 2 first, we multiply 1/4 by 3/4 = 3/16.

But now we have another possibility.

If on the first pick we get an odd number, then we know for the second pick, we need 2 in order to have ODD + EVEN = ODD.

So chance to pick an odd first time around = 3/4. Chance to pick number 2 on the second pick is 1/4. So again we have 3/4 * 1/4 = 3/16. When either situation gives us the desired result, we add the results. so 3/16 + 3/16 = 6/16...reduce to 3/8 for the final answer.

wininblue wrote:
Sorry, I Still don't get it.
Since for the sum to be Odd, one number has to be even, in this case it is 2.
How can the order still matter in this case?

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Re: M10 Q#14 [#permalink] New post 30 Aug 2010, 08:01
Set is : 2,3,5,7

Now: total number of selections = 4X4 [4 ways for the 1st pick & 4 ways for 2nd pick] = 16 = total probable outcomes

total number of favorable outcomes i.e odd sum = selecting 2 in 1st pick and 3or5or7 in 2nd pick. + selecting 3or 5or 7 in 1st pick and 2 in second pick = 1C1X3C1+3C1X1C1 = 3+3 =6

=> P(odd) = 6/16 = 3/8.

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Re: M10 Q#14 [#permalink] New post 05 Jul 2012, 03:04
question stem says prime integers .....wouldn't we consider -2, -3, -5, -7 as well?
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Re: M10 Q#14 [#permalink] New post 05 Jul 2012, 03:07
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teal wrote:
question stem says prime integers .....wouldn't we consider -2, -3, -5, -7 as well?


Only positive numbers can be primes.

For more check Number Theory chapter of Math Book: math-number-theory-88376.html

Hope it helps.
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Re: M10 Q#14 [#permalink] New post 05 Jul 2012, 04:14
Hi,

I understand that prime numbers are only positive numbers starting smallest prime = 2 and onwards......but the confusion was because the question stem said prime integers rather than prime numbers?
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Re: M10 Q#14 [#permalink] New post 05 Jul 2012, 04:16
Expert's post
teal wrote:
Hi,

I understand that prime numbers are only positive numbers starting smallest prime = 2 and onwards......but the confusion was because the question stem said prime integers rather than prime numbers?


Please go through the link in my previous post.

There is no difference between "prime number" and "prime integer".
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Re: M10 Q#14 [#permalink] New post 05 Jul 2012, 23:48
Prime # <10 2,3,5,7
Probablity of sum being odd=1-probability of sum being even
[Lets remember its with replacement type of question]
Prob. of sum being even= 3*3/4*4 [choosing two numbers from 3,5,7 with replacement)+1/16 (choosing two 2's)=10/16=5/8
prob. of sum being odd = 1-5/8=3/8
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Re: M10 Q#14   [#permalink] 05 Jul 2012, 23:48
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