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**zorro** $\displaystyle \therefore$ Pr(X is odd) = Pr(X=1)+Pr(X=3) = $\displaystyle \frac{1}{3} + \frac{1}{6}$ = $\displaystyle \frac{3}{6}$ = $\displaystyle \frac{1}{2}$.......Is this correct?Yes.

And

Pr(XY is even ) = Pr(XY=2) = $\displaystyle \frac{1}{2} + \frac{1}{2}$ = 1.........Is this correct? No. You should be able to check why this is not correct. The probability that XY is equal to is the P(1,2)+P(2,1). That is equal to 1/6 (from your table above). You should also know, from the fundamentals of probability, that those two answers can NOT add up to 1, since that would mean that the probability of any other XY is 0, which is not true from your table. Go slow, and figure out all the possible combinations of X multiplied by Y that will give you an even number. Your possible values of X are 1, 2 and 3 and your possilbe values of Y are 1, 2 and 3.

Pr(X is odd | Y is odd) = $\displaystyle \frac{Pr(X \cup Y) }{Pr(Y)}$ = $\displaystyle \frac{\frac{1}{3} . \frac{1}{6}}{\frac{1}{2}}$.......Is this correct ?No. The probability that Y is odd is not 1/2. It is P(Y=1,3). That probability is 2/3 - calculate that using the marginal probabilities. You never responded if you knew how to calculate marginals. The top of your quotient isn't correct either. It is the P(X AND Y are odd), not X or Y. If it were X or Y, the probability would be 1 on the top. Which is impossible.