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!