Prove that for any three events $A, B, C, P(ABC) \ge P(A) + P(B) + P(C) − 2$ Hi I'm having trouble understanding this question. It is asking for the probability of the intersection of events $A,B,C$, and stating it is greater than or equal to the sum of the probability of each event, minus $2$. 
From what I understood, events are subsets of the sample space and therefore if we add the probability of each event, we should have $1$. Can you help me?
 A: $$1\geq P(A\cap B)=P(A)+P(B)-P(A\cap B),$$ which says
$$P(A)+P(B)\leq1+P(A\cap B).$$
Now, we can use the last inequality twice:
$$P(A)+P(B)+P(C)\leq1+P(A\cap B)+P(C)\leq2+P(A\cap B\cap C).$$
A: We want to prove
$$P(A \cap B \cap C) \geq P(A) + P(B) + P(C) - 2$$
Where $\cap$ denotes intersection of sets.
We know that for any two sets $A$ and $B$, we have
$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$
Which can be rearranged to obtain
$$P(A \cap B) = P(A) + P(B) - P(A \cup B)$$
And hence, since $P(X) \leq 1$ for all $X$, we get
$$P(A \cap B) \geq P(A) + P(B)-1$$ 
Since we are now subtracting a larger number from the right-hand-side.
We can apply the same logic to see that, therefore,
$$P(A \cap B \cap C) = P((A \cap B)\cap C) \geq P(A \cap B) + P(C) - 1$$
Applying the same logic to the right hand side term $P(A \cap B)$ then proves the result
$$P(A \cap B \cap C) \geq P(A \cap B) + P(C) - 1 \geq P(A)+ P(B) + P(C) -2$$
A: Actually it's a special case of the more general inequality
$P(\prod\limits_{i=1}^{n}A_i) \geq \sum\limits_{I=1}^n P(A_i) - (n-1)$
which can be proved by induction on the number of events $n$:
Basis:  n=1, the inequality is trivially true.
Induction hypothesis:: assume that $P(\prod\limits_{i=1}^{n}A_i) \geq \sum\limits_{I=1}^n P(A_i) - (n-1)$, $\forall{n \geq m}$, with $n,\; m \in \mathbb{N}$.
Induction step:
$P(\prod\limits_{i=1}^{m+1}A_i)$
$=P(EA_{m+1})$, where the event $E=\prod\limits_{i=1}^{m}A_i$
$=P(E)+P(A_{m+1})-P(E \cup A_{m+1})$
$\geq P(E) + P(A_{m+1})-1$
$=P(\prod\limits_{i=1}^{m}A_i) + P(A_{m+1})-1$
$\geq \sum\limits_{I=1}^m P(A_i) - (m-1)+P(A_{m+1})-1$, by induction hypothesis
$=\sum\limits_{I=1}^{m+1}P(A_i)-m$.
Hence, we have,
$P(\prod\limits_{i=1}^{n}A_i) \geq \sum\limits_{I=1}^n P(A_i) - (n-1)$, $\forall{n \in \mathbb{N}}$.
The inequality in the question is a special case for $n=3$, let the events be $A_1=A$, $A_2=B$ and $A_3=C$.
A: If you have a partition of the sample space, then they sum to one. 
Note that not all subset are disjoint. 
Hint:
\begin{align}
P(ABC) &= 1- P((ABC)^c)\\
&= 1 - P(A^c \cup B^c \cup C^c)
\end{align}
