Math Probability Question (help me please!)?

I'm opposed to just giving an answer. That just seems like cheating and doesn't help you learn. The thing you want is to know *how* to solve this problem so you can solve similar conditional probability problems yourself.

There are 11 possible rolls where at least one of the rolls is a 3:
3+1 = 4*
3+2 = 5
3+3 = 6*
3+4 = 7
3+5 = 8*
3+6 = 9
1+3 = 4*
2+3 = 5
4+3 = 7
5+3 = 8*
6+3 = 9

Note: be careful not to double-count the 3+3 roll. There are only 11 ways, not 12.

Of these, 5 of them end up with an even sum (see the ones with the asterisks above).

So the probability of an even sum, given that at least one of the rolls is a 3 is 5/11.

Answer:
5/11

P.S. It's easy to make a mistake and think "well if one of the dice is a 3, then the other die has to be a 1, 3 or 5 and the probability of that is 3/6 or 1/2". But that is faulty reasoning and gets you the wrong answer.

5/11

You don't say how many dice you are rolling--it could be 1, 2 or 3, presuming you are rolling normal 6 sided dice. In fact, you don't even say you are rolling dice, although that's what you usually roll to get numbers.

You should have enough in the way of hints or answers to complete the solution. I just wanted to point out that you left out some important information.

I just count. There are 11 rolls that include a 3:
1-3, 3-1, 2-3, 3-2, 3-3, 4-3, 3-4, 5-3, 3-5, 6-3 and 3-6.

"Given that at least one 3 is rolled" means that one of those 11 outcomes occurred. Each is equally likely, and of those 11 outcomes, there are 6 pairs with an even sum.

The probability of an even sum, given that at least one 3 is rolled, is 6/11.

rolls of what?

If it's a single die, the probability that the 2nd die comes up odd is 50%, so the probability that the 2 dice add up to an even number is 50%.

There are 36 possible sums for the roll of two dice. You have to add all the probabilities of the even numbers that are greater than three (4, 6, 8, 10, 12)