If a is to be chosen at random from the integers...

[BTGmoderatorLU]

If a is to be chosen at random from the integers between 1 to 5, inclusive, and b is to be chosen at random from the integers between 6 and 10, inclusive, what is the probability that a + b will be even?

(A) 6/25
(B) 2/5
(C) 12/25
(D) 3/5
(E) 16/25

The OA is C.

I'm really confused with this PS question. Experts, any suggestion? Thanks in advance.

[elias.latour.apex]

For a+b to be even, either both a and b must be even or both a and b must be odd.

Since we will be choosing 5 possible "a" numbers and 5 possible "b" numbers, the total number of possibilities will be 25. The question is: How many of those will fit the above criteria?

We have 3 ways that "a" can be odd and 2 ways that "b" can be odd. So that is 6 possible ways (3x2).
We have 2 ways that "a" can be even and 3 ways that "b" can be even. So that is 6 possible ways.

So, all told, we have 12 possible successes out of 25 possible choices. So the probability will be 12/25.

[Scott@TargetTestPrep]

If a is to be chosen at random from the integers between 1 to 5, inclusive, and b is to be chosen at random from the integers between 6 and 10, inclusive, what is the probability that a + b will be even?

(A) 6/25
(B) 2/5
(C) 12/25
(D) 3/5
(E) 16/25

In order for a + b to be even, we must have a and b both odd or both even.

P(a and b are both odd) = 3/5 x 2/5 = 6/25

P(a and b are both even) = 2/5 x 3/5 = 6/25

So P(a + b = even) = 6/25 + 6/25 =12/25.

Answer: C