A good one

DBushkalov wrote:hi fellows,

here's one task for the community. I only want to know what i do wrong, but i will post a more in depth description of the prompt for those who want to solve it by themselves:

"If X is to be chosen at random fromthe set {1;2;3;4) and Y is to be chosen at random from the set {5;6;7}, what is the probability that X*Y will be even?"

A) 1/6
B) 1/3
C) 1/2
D) 2/3
E) 5/6

Here is how the GMAC guys calculate it:

there are 12 (4*3) possible products of X*Y where X is chosen from the first set and Y is chosen form the second set. The possible outcomes for 2 odd numbers are 4 (2*2) --> 4/12 is the probability that the product will be odd, thus 8/12 is the probability that the product will be EVEN.

Here is how I calculated it:

if we get an even number from the first set OR, i repeat, OR, an even number from the second set --> VIOLA ! we have an even number when X*Y. Lets now calculate the probability that we draw out an even number from the first and second sets:
1-st set: there are 2 even numbers from 4 ---> 2/4 = 1/2 is the probability of drawing an even nr. from the first set.

2-nd set: there is 1 even number ---> 1/3 is the probability to draw it out

From 1 and 2 follows that the total probability for an even number to be drawn is = 1/2 + 1/3, which yields 5/6.

5/6 is not 2/3 ,though. Can you please tell me what i do wrong ?

Thank you in advance