The solution pasted below was provided by Knewton Instructor Rich. It's quite clear.

Notice that the set has 4 numbers and each statement involves combinations of 2. How many possible combinations of 2 can be made out of set of 4?

In general, if you have a set of n items and you're trying to find a subgroup of k items, the number of combinations is:

n! / [k! * (n-k)! ]

So for this problem, since we have a set of 4, and we're interested in subgroups of 2, there are 4! / (2!*2!) = 6 combinations.

Now, notice from Statement 1 involves a probability of 5/6. That means exactly 5 out of the 6 combinations have a product that is even. Which also means only 1 out of 6 combinations has a product that is odd. The only way we can multiply two integers and get an odd product is if both integers are odd. That means only one pair of odd values are in the set, which therefore means only two of the numbers are odd and the other two must be even. Sufficient.

Same thing for Statement 2. We know that the probability that the two numbers sum to an even integer is 1/3. That means that exactly 2 out of the 6 combinations sum to an even integer. The only way to get an even sum from two integers is if either both integers are odd or both are even. Try mapping out the possibilities, and you'll see that the only way you can get this exact condition is if 2 of the numbers are odd and the other 2 are even. Sufficient.

karthikpandian19 wrote:Set G contains 4 distinct positive integers. How many of the integers are even?

1. The probability that the product of two integers randomly selected from set G is even is 5/6.
2. The probability that the sum of two integers randomly selected from set G is even is 1/3.

It is almost always easier to check for odd product and odd sum, since they happen only under a distinct set of circumstances.

We have 4 distinct positive integers in the set. We need to see if we can find how many are even.

1. The probability that the product of two integers randomly selected from set G is even is 5/6.

p(even) = 5/6 (given). Hence p(odd) = 1-5/6 = 1/6. We only get odd product when both numbers are odd. Since probability of odd product is not 0, at least 2 of them are odd. Let's check 2 odd 2 even case. Probability = 2C2/4C2 = 1/6. Success. Hence we know that there are 2 odds and 2 evens. Sufficient.

2. The probability that the sum of two integers randomly selected from set G is even is 1/3.
Again:
p(even sum) = 1/3 => p(odd sum)= 2/3. This means that there is at least one odd number.
Since we know from 1st statement that there were 2 odds, let's check the 2 odds case.

p (odd sum) = 2C1*2C1/(4C2) = 2*2*2/(4*3) = 2/3.
Hence we know that there are 2 odds and 2 evens. Success. Sufficient Statement.

D is correct.