Maths Probability Question...please help!?

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.

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 cant really gove you an answer,but what I can give you is a way to a solution, that is you have to find the anglde that you relate to or peaks your interest. A good paper is one that people get drawn into because it reaches them ln some way.As for me WW11 to me, I think of the holocaust and the effect it had on the survivors, their families and those who stood by and did nothing until it was too late.