Prove that for any three events $A, B, C, P(ABC) \ge P(A) + P(B) + P(C) − 2$

Question

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?

Answer 1

$$1\geq P(A\cap B)=P(A)+P(B)-P(A\cap B),$$ which says $$P(A)+P(B)\leq1+P(A\cap B).$$ Now, we can use the last inequality twice: $$P(A)+P(B)+P(C)\leq1+P(A\cap B)+P(C)\leq2+P(A\cap B\cap C).$$

Answer 2

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}

Answer 3

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

$$P(A \cap B \cap C) \geq P(A \cap B) + P(C) - 1 \geq P(A)+ P(B) + P(C) -2$$

Answer 4

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 $A_3=C$.