I have this problem that states:
$A = \{\text{red die is}~3, 4,~\text{or}~5\}$
$B = \{\text{blue die is}~1~\text{or}~2\}$
$C = \{\text{sum is}~7\}$
Both the red and blue die are under equally likely probability.
I need help finding if they are pairwise independent and if they are mutually independent. The problem is I don't quite fully understand what those two terms mean. I read the definition and examples on Wikipedia but there's so much terminology on there that makes no sense to me.
From my understanding, they are not pairwise independent because if you do $A \cap B \cap C$, $A$ doesn't have any intersection with $B$. They do intersect with $C$ if you roll a 5 on the red die and a 2 on the blue die. But they don't all share something.
For mutual independence, I can't find any definition that makes sense to me.
As you can see I'm not that great at Discrete Mathematics, so if you can explain in easy terms it would be much appreciated.
Best Answer
$A$ and $B$ are independent because the red and blue dice are separate dice. (If one embedded magnets in them and threw them next to each other one could question such things, but it is conventional in phrasing math problems like this to construe that as meaning they will be independent.)
Now consider $\Pr(C \mid A\ \&\ B)$ and $\Pr(C)$. If these are equal then $C$ is independent of $[A\ \&\ B]$ but that does not mean that $C$ is independent of $A$, nor that $C$ is independent of $B$. What is the conditional distribution of the outcomes given by the red and blue dice given the event $[A\ \&\ B]$? Given that event, six outcomes are equally probable: $$ \begin{array}{c|ccc} & 3 & 4 & 5 \\ \hline 1 & 1/6 & 1/6 & 1/6 \\ 2 & 1/6 & 1/6 & 1/6 \end{array} $$ In only one of these outcomes, the sum is $7$; therefore $\Pr(C\mid A\ \&\ B) = 1/6$.
What is the marginal (i.e. unconditional) probability of $C$? It is \begin{align} & \Pr\Big( \Big[ 1\ \&\ 6\Big] \text{ or } \Big[ 2\ \&\ 5\Big] \text{ or } \Big[ 3\ \&\ 4\Big] \text{ or } \Big[ 4\ \&\ 3\Big] \text{ or } \Big[ 5\ \&\ 2\Big] \text{ or } \Big[ 6\ \&\ 1\Big] \text{ or } \Big) \\[10pt] = {} & \frac 1 {36} + \frac 1 {36} + \frac 1 {36} + \frac 1 {36} + \frac 1 {36} + \frac 1 {36} = \frac 1 6. \end{align} Hence we have $\Pr(C) = \Pr(C\mid A\ \&\ B)$, so $C$ is independent of $[A\ \&\ B]$.
Is $C$ independent of $A$? It is if $\Pr(C\mid A) = 1/6$; otherwise it's not.
\begin{align} & \Pr(C\mid A) = \Pr(\text{sum is 7} \mid \text{red is 3, 4, or 5} ) = \frac{\Pr(\text{sum is 7 and $A$ is 3, 4, or 5})}{\Pr(\text{red is 3, 4, or 5})} \\[10pt] = {} & \frac{1/36 + 1/36 + 1/36}{1/2} = \frac 1 6. \end{align} Therefore $A$ and $C$ are independent.
Similarly \begin{align} & \Pr(C\mid B) = \Pr(\text{sum is 7} \mid \text{blue is 1 or 2} ) = \frac{\Pr(\text{sum is 7 and blue is 1 or 2})}{\Pr(\text{blue is 1 or 2})} \\[10pt] = {} & \frac{1/36+1/36}{1/3} = \frac 1 6. \end{align} Therefore $B$ and $C$ are independent.
Thus $A$, $B$, and $C$ are pairwise independent.
Now \begin{align} \Pr(A\ \&\ B\ \&\ C) = \Pr(A)\Pr(B)\Pr(C\mid A\ \&\ B) = \frac 1 2 \cdot \frac 1 3 \cdot\frac 1 6 = \frac 1 {36} = \Pr(A)\Pr(B)\Pr(C) \end{align} and so the three are mutually independent.