Compute the probability of each of the following events

Question : Two discrete random variables X and Y have joint pmf given by the following table:

$\displaystyle

\begin{array}{c|c|c|c}.& Y=1 & Y=2 & Y=3 \\

\hline

X=1 & \frac{1}{12} & \frac{1}{6} & C \\

X=2 & \frac{1}{6} & \frac{1}{4} & \frac{1}{12} \\

X=3 & \frac{1}{12} & \frac{1}{12} & 0 \\

\end{array}$

i) Find C

ii) Compute the probability of each of the following events.

- X $\displaystyle \le $ 1 1/2

- X is odd

- XY is even

- Y is odd given that X is odd

I am still having problems

Quote:

Originally Posted by

**ANDS!** Remember that the probability that an outcome occurs in a sample space is 1 (or 100% chance that some event is going to happen). So if you know that the Probability of X and Y occurring MUST add up to 1, and you have all probabilities except for C, how do you go about figuring out what C is?

Here is what u have done till now....

C = $\displaystyle \frac{1}{12}$ ..........Is that correct?

Could u please explain me what does 'X is odd' mean, i tried searching on the net for the term odd random variable but couldnt find any meaning ...........please explain me when is a random variable termed odd or even