Question

HELLLP!! Ill give medals..I'm working with probabilities!!



Two number cubes, a red one and a blue one, are rolled. What is the probability that the red cube will be an odd number while the sum of the two dice is an even number?

Answer 1

There are 4 cases on the sum of numbers
1.)odd+odd=even
2.)odd+even=odd
3.)even+odd=odd
4.)even+even=even
now it is given that the sum is even
thus we can remove case 2 and 3
the cases we have now are
1.)odd(red)+odd(blue)=even
2.)even(red)+even(blue)=even
clearly probability that red will be odd is \[\frac{1}{2}\]
Although I'm not sure myself

Answer 2

I'll call someone to help you, @Kainui

Answer 3

\[P(reddodd|sumeven)=\frac{P(redodd-sumeven )}{P(sumeven))}\]
here by - I mean intersection, probability of red odd given that sum is even is given by probability of red odd and sum even divided by probabilityof sum even
P(redodd-sumeven)=\[\frac{1}{4}\], which is probability of sum is even and red is odd
\[P(sumeven)=\frac{2}{4}=\frac{1}{2}\] which is probability of sum is even
\[P(redodd|sumeven)=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{4} \times 2=\frac{1}{2}\]

Answer 4

So I think it's \[\frac{1}{2}\] but you should confirm from somebody else

Answer 5

@k_lynn @iGreen.

Answer 6

"What is the probability that the red cube will be an odd number \(while\) the sum of the two dice \(is\) an even number?"
I find the question not very explicit. The fact that the sum \(is\) an even number (and not "will be") seems to imply that it is a condition, in which case the conditional probability calculations done by @Nishant_Garg is good.
There is a slight chance that the \(while\) is meant to be a joint condition, in which case we only have 1/4 of the possible cases. I.e. \(P(red~odd \cap sum~even)\)