I have difficulty in probability problem

[bia]

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?
A. 1/8
B. 1/6
C. 3/8
D. 1/2
E. 5/8

[mim3]

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?
A. 1/8
B. 1/6
C. 3/8
D. 1/2
E. 5/8

Is the OA C?

I had a little trouble with this one too- here's how I got C:

S: 2, 3, 5, 7

The only way that the sum of two integers from the set is odd is if it's 2+3,5 or 7. So, two "winning" scenarios

So 2 can be either the 1st number selected or the 2nd number...
Prob that 2 is the 1st number selected:
(1/4)
Prob that 2 is NOT the second number is:
(3/4)

(1/4)(3/4)= 3/16

On the flip side:
Prob that the first number is not 2:
(3/4)
Prob that the second number IS 2:
(1/4)

(1/4)(3/4)= 3/16

Two winning scenario prob added together:
6/16= 3/8

Again, not confident this is the correct solution. I'd be interested to hear other approaches...

[chidcguy]

Total number of ways of selecting two numbers is 16. First number 4 ways and 2nd number 4 ways.

The only way we can get an odd number is if we add one odd and one even.

The only even is 2 and hence 2 must be a member of the pair. 2 can be the first or second but has to be in there. The total number of such pairs would be 6 ie (2,3) (2,5) (2,7) and (3,2) (3,5) (3,7)

Probability is 6/16 which is 3/8

[gmatinjuly]

I agree answer is 3/8
Set S is as below
S = {2 3 5 7}

Probability of getting an odd sum is when

first even number then pick any odd + First odd then pick any even
¼ * ¾ + ¾ * ¼
= 3/16 + 3/16
= 3/8