As you pick the first number, it doesn't really matter which one you pick. The only thing the first number changes is what options you have when you pick the second number.

Two Possible Situations: 1st: Pick 2 and then forst 2nd you'd have to pick one of the odd numbers in order to get EVEN + ODD = ODD.

So, Chance of picking 2 on the first try? 1 out of 4 = \(\frac{1}{4}\)

Now on the second pick, we must have one of the odd numbers 3, 5, or 7, so 3 out of 4, or \(\frac{3}{4}\). In order to figure out the chance of getting an odd when we pick 2 first, we multiply 1/4 by 3/4 = 3/16.

But now we have another possibility.

If on the first pick we get an odd number, then we know for the second pick, we need 2 in order to have ODD + EVEN = ODD.

So chance to pick an odd first time around = 3/4. Chance to pick number 2 on the second pick is 1/4. So again we have 3/4 * 1/4 = 3/16. When either situation gives us the desired result, we add the results. so 3/16 + 3/16 = 6/16...reduce to 3/8 for the final answer.

wininblue wrote:

Sorry, I Still don't get it.

Since for the sum to be Odd, one number has to be even, in this case it is 2.

How can the order still matter in this case?

Posted from my mobile device

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J Allen Morris

**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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