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Boole's Inequality

Probability is the possible chance of happening of an event. It lies between 0 and 1. Boole’s inequality is a concept of probability theory in mathematics. It is also known as a union bound. This theory was named after George Boole.

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According to the Boole’s inequality, given the countable or finite set of events, the probability of happening of at least one of the events is less than or equal to the sum of the probabilities of the events taken individually. Mathematically we can represent it as:
$P\ (\cup_i\ [E_i])\ \leq\sum_i\ (P\ (E_i))$
Where, $E_i$ are the events

Boole’s Inequality Proof

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Boole’s Inequality Proof

According to the Boole’s inequality,

$P\ (\cup_i\ [Y_i])\ \leq\ \sum_i\ (P\ (Y_i))$

If $X_1$, $X_2$, $….$, are disjoint subsets in the probability space given, then,

$P\ (\cup_i\ (X_i))$ = $\sum_i\ (P\ (X_i))$

This is also known as countable additivity.

If we have two subsets such that $X \subset Y$, then $P\ (X)\ \leq\ P\ (Y)$.

We also have, $P\ (X)$ = $P\ (Y)\ –\ P\ (Y\ –\ X)\ \geq\ 0$

Now suppose,

$X_i$ = $Y_i\ -\ \cup_{(j\ =\ 1)}^{(i – 1)}\ (Y_j)$

This will give us $X_i\ \subset\ Y_i$ for all $i$. Also, since $X_i$ are disjoint, we have,

$\cup_{(i\ =\ 1)}^\infty (X_i)$ = $\cup_{(i\ =\ 1)}^\infty (Y_i)$

$\Rightarrow$ $P\ (\cup_i\ (Y_i))$ = $P\ (\cup_i\ (X_i))$

$\Rightarrow$ $P\ (\cup_i\ (Y_i))$ = $\sum_i\ [P\ (X_i)]$

$\Rightarrow$ $P\ (\cup_i\ (Y_i))\ \leq\ \sum_i\ [P(Y_i)]$

Hence, proved.

Boole’s Inequality Proof by Induction

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We prove Boole’s inequality by the principal of mathematical induction.

According to the Boole’s inequality,

$P\ (\cup_i\ [E_i])\ \leq\ \sum_i\ (P\ (E_i))$

For $n$ = $1$, we have

$P\ (E_1)\ \leq\ P\ (E_1)$

This is true in all cases.

Let us assume that the result holds true for $n$ = $k$

$\Rightarrow$ $P\ (\cup_{(i\ =\ 1)}^k\ [E_i])\ \leq\ \sum_{(i\ =\ 1)}^k\ (P\ E_i))$

We will prove that the result is also true for $n$ = $k\ +\ 1$


$P\ (X\ \cup\ Y)$ = $P\ (X)\ +\ P\ (Y)\ –\ P\ (X\ \cap\ Y)$

So we have,

$P\ (\cup_{(i\ =\ 1)}^{(k\ +\ 1)}\ [E_i])$ = $P\ (\cup_{(i\ =\ 1)}^k\ [E_i])\ +\ P\ (E_{(k\ +\ 1)})\ –\ P\ ((\cup_{(i\ =\ 1)}^k\ [E_i])\ \cap\ E_{(k\ +\ 1)})$

We know that,

$P\ ((\cup_{(i\ =\ 1)}^{k}\ [E_i]) \cap\ E_{(k\ +\ 1)})\ \geq\ 0$

Therefore, we have

$P\ (\cup_{(i\ =\ 1)}^{(k\ +\ 1)}\ [E_i])\ \leq\ P\ (\cup_{(i\ =\ 1)}^k\ [E_i])\ +\ P\ (E_{(k\ +\ 1)})$

$\Rightarrow$ $P\ (\cup_{(i\ =\ 1)}^{(k\ +\ 1)}\ [E_i])\ \leq\ \sum_{(i\ =\ 1)}^k\ (P\ (E_i))\ +\ P\ (E_{(k\ +\ 1)})$

$\Rightarrow$ $P\ (\cup_{(i\ =\ 1)}^{(k\ +\ 1)}\ [E_i])\ \leq\ \sum_{(i\ =\ 1)}^{(k\ +\ 1)}\ (P\ (E_i))$

Since is result is proven to be true for $n$ = $k\ +\ 1$, when it is true for $n$ = $1$ and $n$ = $k$, therefore, the results holds true for all $n$.

Boole’s Inequality Probability Theory

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The Boole’s inequality or the union bond is applicable at places when we have to show that the union probability of some events is less than a particular value. It is a simply yet a very useful result. It is used frequently in multiple types of applications.


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Let us see some examples that make use of Boole’s inequality.
Example 1: Supposedly we have to throw m distinguishable balls in m distinguishable boxes that too randomly such that a box can contain balls in any number may even 0 balls. 

Solution: Consider an event that box 2 is empty. This will mean ball $j$ is not in box 2. This can now be proceeding taking up events and using Boole’s inequality.

Example 2: Suppose we are given that for a certain random variable say $X$, $E\ (X)$ = $4$. Find the largest value for $P\ (X\ \geq\ 50)$.

Solution: We know that $X$ is non negative that is $X\ \geq\ 0$. 

$\Rightarrow$ $P\ (X\ \geq\ 0)$ = $1$.

Therefore, the probability distribution for the variable X lies in the interval $[0,\ \infty)$.

When we balance the distribution at mass point of 4, with all of the mass lying in the given interval above, we cannot get much mass in the interval $[50,\ \infty)$. The distribution will balance at point 4 only when there is a proportion of $\frac{4}{50}$ at $50$. This will imply that $P\ (X\ \geq\ 50)$ can have maximum value of  $\frac{4}{50}$. Mathematically we can prove it as below.

$\sum_{(for\ all\ x)}\ [x\ P\ (X$ = $x)]$ = $4$

The terms where $x\ \geq\ 50$, will contribute,

$\sum_{(x\ \geq\ 50)}\ [x\ P\ (X$ = $x)]\ \geq\ \sum_{(x\ \geq\ 50)}\ [50\ P\ (X$ = $x)]$ = $100\ P\ (X\ \geq\ 50)$

Since $X\ \geq\ 0$ therefore, all the terms are non negative.

$\Rightarrow$ $4\ \geq\ 100\ P\ (X\ \geq\ 50)$ 
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