# Flipping a fair coin three times: How to find independent/dependent events?

Given that I have a fair coin which I toss three times, I have the following sample space: $S=\left\{HHH , HHT , HTH, THH , HTT, TTH, THT , TTT\right\}$

how do I:

a) Find three events A , B, and C such that no two are independent but $P(A\cap B\cap C)=P(A)P(B)P(C)$ ?

b) Find three events A , B, and C such that every two are independent but all three are not independent: $P(A\cap B\cap C)≠P(A)P(B)P(C)$ ?

How would I approach a problem like this ? I started by drawing a binary tree that ended up with 8 leaves but I have no idea how to use it to get the answers. Any help would be appreciated.

You could take a somewhat number-theoretic approach. This is perhaps not the most obvious approach; I wouldn't be surprised if you get another answer that uses more probabilistic insight; but it's fun and it works.

There are $8$ equiprobable elementary outcomes, so the probability of an event $E$ has the form $k_E/8$ with $0\le k_E\le8$ an integer. Thus, for a), we need to have

$$\frac{k_{ABC}}8=\frac{k_A}8\cdot\frac{k_B}8\cdot\frac{k_C}8\;,$$

or

$$2^6k_{ABC}=k_Ak_Bk_C\;.$$

So there must be a total of six factors of $2$ in the three $k_E$. But we can't have three factors of $2$ in any of them, since then it would be the certain event, which is independent of all events. Hence each of them has exactly two factors of $2$. That doesn't leave any room for any other factors, so we must have $k_{ABC}=1$ and $k_A=k_B=k_C=4$; that is, the intersection must contain a single outcome, and each of the three events must have probability $1/2$.

That reduces our search space quite considerably. Let's forget about the binary structure of the process and view the eight outcomes simply as eight different atomic events, labeled by $1$ through $8$. Then the triple intersection can be any one of them, say, $P(A\cap B\cap C)=\{1\}$. The event $A$ can contain any four of them, say, $A=\{1,2,3,4\}$. Then we just have to choose the other two such that they both contain $1$ and don't intersect $A$ or each other in exactly two outcomes. That's rather straightforward to do; one possible solution is $B=\{1,2,3,5\}$ and $C=\{1,6,7,8\}$.

For b), the events must be pairwise independent, so e.g.

$$\frac{k_{AB}}8=\frac{k_A}8\cdot\frac{k_B}8,$$

or

$$2^3k_{AB}=k_Ak_B\;.$$

Again we can't have three factors of $2$ in one $k_E$, as that would make all three events independent. Thus both must be even, and at least one of them must be $4$. Since this is true for all three pairs, at least two, say, $k_A$ and $k_B$, must be $4$, and the third, $k_C$, must be even.

For $k_C=2m$, the triple product is $m/16$, which for odd $m$ can't be the probability of the triple intersection, so we don't have to worry about the triple intersection in those cases. The case $m=1$ seems easiest to deal with. So let $C=\{1,2\}$, and then we need to find the other two such that they each intersect $C$ in one outcome and intersect each other in two outcomes. This is again rather straightforward to do; one possible solution is $A=\{1,3,4,5\}$ and $B=\{2,3,4,6\}$, with the triple intersection "too small" (namely empty), and another is $A=\{1,3,4,5\}$ and $B=\{1,3,6,7\}$, with the triple intersection "too big" (namely $\{1\}$).