When you roll 2 dice there are 36 equally likely options, starting at double one, $(1,1)$, and going all the way to double six, $(6,6)$. Of these, there are 11 equally likely ways to roll a $5$, $(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3),(5,4),(5,6)$, so the odds of rolling a 5 is $\frac{11}{36}$.
Similarly there are 18 equally likely ways to roll an odd number (I'll let you think about which combinations these are), so the odds of rolling an odd number are $\frac{18}{36}=\frac{1}{2}$.
At this point you may think the odds of rolling a $5$ or and odd number should be $\frac{11}{36}+\frac{18}{36}=\frac{29}{36}$, however this is not the case. Note that six of the rolls contain both a $5$ and are odd:
$(2,5), (4,5), (6,5), (5,2), (5,4),(5,6)$
If we just add the two probabilities, we'll count these situations twice, even though they're no more likely to occur than any other combination! In reality there are only $11+18-6=23$ dice rolls which contain a $5$ or are odd (see if you can list them), and so the odds of rolling either a $5$ or an odd combination is $\frac{11}{36}+\frac{18}{36}-\frac{6}{36}=\frac{23}{36}$.
This is a simple application of the "principle of inclusion and exclusion", which you may want to look up.
$Cov(Y, Z)=Cov(X_1+X_2,\,4X_1-X_2)$
$=4Cov(X_1,X_1) -Cov(X_1, X_2)+4Cov(X_2, X_1)-Cov(X_2, X_2)\,\,by\, linearity$
$=4Var(X_1)+3Cov(X_2, X_1)-Var(X_2)$
$=4Var(X_1)+0-Var(X_2) \,\,\,since\, X_1 \,and\, X_2\, are\, independent$
Best Answer
a) Correct.
b) Hint: Independence.
Also $\mathsf E(XY) =\sum\limits_{x=1}^6\sum\limits_{y=1}^6 \dfrac{xy}{36}$