probability: even xy

[Uri]

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

source: GMAT OG- Quant Review
PS prob: 80

can you please explain where the mistake lies in the following logic?

xy will be even if either one of x or y is even or both are even.

probability of choosing even x and any y=1/2
probability of choosing even y and any x= 1/3
probability of choosing both x and y even= 1/6
probability of getting even xy= (1/2) + (1/3) + (1/6)= 1!!!

how would you approach the problem?

[rossmj]

You are double counting sets of numbers. The best way to approach this is to establish the total number of possibilities which in this case is 12 (3 options for 1, 3 options for 2, 3 options for 3, and 3 options for 4). Then look and see how many of these options result in an even number: 1 for 1,3 for 2,1 for 3,&3 for 4 =8. 8/12=2/3.

With your logic you are say 1/2 of the 12 will be even bc x is even, 1/3 of the 12 will be even bc y is even, and 1/6 of the 12 will be even bc x and y are even. If you add this up you indeed get 12 which would equal a probability of 1. But surely some of the 1/3 is contained in the 1/2 and all of the 1/6 are included in the 1/2 and 1/3.

[g2000]

My approach is slightly different.

P(xy is even) = 1 - P(xy is odd)

In order to get odd for xy, both x and y need to be odd.
P(x is odd) = 1/2
P(y is odd) = 2/3
P(xy is odd) = 1/2 * 2/3 = 1/3

P(xy is even) = 1 - P(xy is odd)
= 1 - 1/3
= 2/3

[rossmj]

Nice approach, and certaintly more effective for large sets and more applicable to Uri's original post.

[vertigo05]

what is wrong with this approach?

P(x is even)= 1/2
P(y is even) = 1/3

P(xy is even)=1/2*1/3 = 1/6

[target790]

last approach is not wrong but incomplete.You are ignoring the case (odd*even=even).


S here we have 8 such total cases(even*even,even*odd)

so answer is 8/(4C1*3C1)=2/3