Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 4 / b

Use the ordered outcomes of the two rolls as the sample space:Thus each elementary outcome has probability $1/36$. This is the usual probability space for two independent fair dice.

For an outcome $\omega =(i,j)$, the two random variables are the coordinate mapsTheir marginal probability mass functions are $p_X(i)=p_Y(j)=1/6$ on $\{1,\dots ,6\}$, and their joint probability mass function satisfiesOff that support, the same factorization holds with zeros. Hence $X$ and $Y$ are independent random variables, by the joint probability mass function criterion for independence.

There are three even faces, so $\mathbb{P}(A)=1/2$. The sum is even precisely when both faces have the same parity, giving $9+9=18$ outcomes and $\mathbb{P}(B)=1/2$. Both $A$ and $B$ occur precisely when both faces are even, giving nine outcomes. ThereforeThus $A$ and $B$ are independent events.

No: pairwise independence does not imply that the three are independent events. Take $C=\{Y\text{ is even}\}$. Then $\mathbb{P}(C)=1/2$, and $\mathbb{P}(A\cap C)=9/36=1/4$ by independence of the two rolls. Also $B\cap C$ is precisely the event that both faces are even, so $\mathbb{P}(B\cap C)=1/4$. Thus all three pairs of events are independent.

But $A\cap B\cap C$ also consists of the nine even-even outcomes, soThis violates the triple-intersection requirement for independent events.