# Thread: 30.8 - Events and Probability of dice

1. ## 30.8 - Events and Probability of dice

Three dice are rolled. What is the probability that the sum of the numbers showing is even? Explain.

-I had a quiz with this question and I got it wrong. I assume I had an improper sample space??

Thanks

2. Originally Posted by ccdelia7
Three dice are rolled. What is the probability that the sum of the numbers showing is even? Explain.

-I had a quiz with this question and I got it wrong. I assume I had an improper sample space??

Thanks
Three fair, six-sided dice? The answer should be 1/2, I think.

3. Originally Posted by ccdelia7
Three dice are rolled. What is the probability that the sum of the numbers showing is even? Explain.

-I had a quiz with this question and I got it wrong. I assume I had an improper sample space??

Thanks
The sum is even when you have exactly three even or exactly one even, the number of evens
is binomial $\sim B(3,0.5)$, so:

$p(even\ sum)=b(1;3,0.5)+b(3,3,0.5)=3 \times 0.5^3 + 0.5^3=0.5$

RonL

4. I assume you're right, that the odds are 1/2 because if you think about it,

- Any even number added to any even number added to any even number is even.

- Any number added to any even number added to any odd number is odd

-Any odd number added to any odd number added to any odd number is odd.

- Any odd number added to any odd number added to any even number is even.

Are there any other cases that I'm neglecting?? That's where I have the problem!

5. Originally Posted by ccdelia7
Three dice are rolled. What is the probability that the sum of the numbers showing is even? Explain.
Originally Posted by icemanfan
Three fair, six-sided dice? The answer should be 1/2, I think.
Yes... icemanfan is right
Originally Posted by ccdelia7
-I had a quiz with this question and I got it wrong. I assume I had an improper sample space??
Thanks
What?!!You listed(or even thought of trying to list) all the 108 possible throws that gave an even sum?

To prove the answer is 0.5, consider the fact that only throws which have all even numbers and the ones with 1 odd number gives a even sum...

EDIT: Yes, you are right. I was trying to explain the very thing you wrote...What did you do in the quiz??

6. ## thanks

No i was just trying to think of the probability logically, and it seemed as though the chances would be higher than 1/2, like 3/2. But i see now, what the real approach should be.

Thanks to everyone!