Let $(A,B,C)$ be a probability space, and consider events $F$, $G$, and $H$ for which $P(F) \gt P(G) \gt P(H) \gt 0$. Events $F$ and $G$ form a partition of $A$, and events $F$ and $ H$ are independent. Can events $G$ and $H$ be disjoint?
Although there is a Hint: Suppose $G$ and $H$ are disjoint, then you will get a contradiction. However, I still have no idea about this, can anyone give me some hint a little more?
On
The condition that $F$ and $G$ form a partition of the space $A$ means that $F\cup G=A$ and $F\cap G=\varnothing$. If we suppose that $G$ and $H$ are disjoint, then necessarily $H\subseteq F$. In fact, $H\subsetneq F$ — they can't be equal because $P(F)>P(H)$. Now state the definition of independence of two events, and see that this equation (that defines independence) cannot hold true — that will be the desired contradiction.
$F$ and $G$ form a partition of $A$ tells us two things, namely
Furthermore, the sample space being $A$ tells us that $H \subseteq (A = F \cup G)$. Clearly, $H \subseteq F$ and/or $H \subseteq G$.
Furthermore, if we consider $P(F) \gt P(G) \gt P(H) \gt 0$, then $H \subset F$ and/or $H \subset G$.
Suppose $G$ and $H$ are disjoint. Then $(H \not\subset G) \rightarrow (H\subset F)$ or in other words, $H \cup F = H$
By the inclusion exclusion principle, $|H \cup F| = |H| = |H| + |F| - |H \cap F| \iff |F| = |H \cap F|$,
which is a contradiction since $P(F) > P(H) \geq P(F \cap H) \iff |F| > |H \cap F|$ .