Seems like a copy. Please check the earlier post.
https://gre.myprepclub.com/forum/x-is-chos ... 24517.html

Quantity A: P(y is any number)=1 and P(x is even)=1/2, hence P(xy is even)=1/2
Quantity B: P(y is always odd)=1 and P(x is odd)=1/2, hence P(x+y is even)=1/2

answer is C

Given that x is chosen at random from the set {1,2,3,4} and y is chosen at random from the set {5,7,9}.

Quantity A: The probability that xy will be even

For xy to be even at least one of them should be even.
But we can see that y can take only odd values
=> For xy to be even x has to even
=> P(xy = Even) = P(x = even) = \(\frac{2}{4}\) = \(\frac{1}{2}\)

(Because for x we have two choices (2,4) to get even out of 4)

Quantity B: The probability that (x+y) will be even

For x + y to be even, either both should be odd or both should be even, but y cannot be even
=> for x + y = even, both x and y should be odd. We know that y is always odd, so we just need to find the cases in which x is odd too.

=> P(x+y = even) = P(x=odd)= \(\frac{2}{4}\) = \(\frac{1}{2}\)

Clearly, Quantity A(\(\frac{1}{2}\)) = Quantity B(\(\frac{1}{2}\))

So, Answer will be C.
Hope it helps!

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