Boole’s Inequality

Boole's Inequality

Boole's Inequality

The Boole's Inequality Theorem states that "the probability of several events occuring is less than or equal to the sum of the probabilities of each event occuring".

For any two events $A$ and $B$, we have

 $$\begin{eqnarray*} P(A \cup B) &=& P(A) + P(B) - P(A\cap B)\\ &\leq & P(A) + P(B)\\ & & \quad (\because P(A\cap B)\geq 0) \end{eqnarray*}$$

Similarly, for three events $A$, $B$ and $C$, we have

 $$\begin{equation*} P(A \cup B\cup C) \leq P(A) + P(B) +P(C). \end{equation*}$$

Boole's Inequality Theorem

For $n$ events $A_1,A_2,\cdots, A_n$

 $$$$\label{bool} P\big(\cup_{i=1}^n A_i\big)\leq \sum_{i=1}^n P(A_i).$$$$

Proof

For any two events $A_1$ and $A_2$,

 $$$$\label{bool01} P(A_1\cup A_2) = P(A_1) + P(A_2) - P(A_1\cap A_2)\leq P(A_1)+P(A_2)$$$$

Hence \eqref{bool} is true for $n=2$.

Suppose \eqref{bool} is true for $n=r$. That is

 $$$$\label{bool2} P\big(\cup_{i=1}^r A_i\big)\leq \sum_{i=1}^r P(A_i).$$$$

For $n=r+1$,

 $$\begin{eqnarray*} P\big(\cup_{i=1}^{r+1} A_i\big) & = & P\big(\cup_{i=1}^{r} A_i \cup A_{r+1}\big)\\. &\leq & P\big(\cup_{i=1}^{r} A_i\big) + P(A_{r+1}) \text{ (Using \eqref{bool01})}\\ & \leq & \sum_{i=1}^r P(A_i) + P(A_{r+1})\text{ (Using \eqref{bool2})}\\ &\leq & \sum_{i=1}^{r+1} P(A_i). \end{eqnarray*}$$

Hence \eqref{bool} is true for $n = r+1$. Thus by the Principle of mathematical induction, \eqref{bool} is true for all n.

Conclusion

In this tutorial, you learned about Boole's Inequality.

To read more about the tutorials on Probability Theory refer the link Probability Theory. These tutorials will help you to understand basic concepts of probability and various important results of probability theory along with some numerical solved examples on probability theory.