If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?

(A) 1/6

(B) 1/3

(C) 1/2

(D) 2/3

(E) 5/6

If x is to be chosen at random from the set {1, 2, 3, 4} and y is to be chosen at random from the set {5, 6, 7}, what is the probability that xy will be even?

(A) 1/6

(B) 1/3

(C) 1/2

(D) 2/3

(E) 5/6

For the product xy to be even, you need to pick an even number for at least one of the 2 values, x or y. So let’s see how many possibilities we can get

- Picking 2 from the first set (2, 5), (2,6), (2,7) → 3
- Picking 4 is also similar, so → 3
- Picking 6 from the second set (1,6), (3,6) → 2

Note: When we pick 6 from the second set in the third set of possibilities, we are ignoring 2 and 4 because we’ve already considered (2,6) and (4,6) in the above cases.

So we have a total of **8** possibilities to get an even xy. How many total ways to pick xy? it’s just 4 (elements in set 1) times 3 (elements in set 2) = **12**.

So the probability of getting even xy is 8/12 = 2/3

I arrived at the same answer (2/3) but from a different approach. Yours seems like an easier way but would it be possible to please check the logic if it seems correct?

[E stands for Even and O stands for Odd]

P(X-E) = P(X-O) = 1/2

P(Y-E) = 1/3

P(Y-O) = 2/3

For product xy to be even, either of the following cases should be true:

- x & y are even

In which case, P(X and Y Even) = 1/2 multiplied by 1/3 = 1/6 - x is even and y is odd

In which case, P(X - E and Y - O) = 1/2 multiplied by 2/3 = 1/3 - x is odd and y is even

In which case, P(X - O and Y - E) = 1/2 multiplied by 1/3 = 1/6

Since these are mutually exclusive probabilities, adding the three probabilities = 4/6 = 2/3

It would be a great help for similar sums in the future.

Perfect! Your solution is actually a better mathematical approach than mine.

Thank you!