Probability of the Union

3. Probability: Probability

Probability of the Union

Now that it is known what the probability of the intersection of $A$ and $B$ looks like, it is possible to calculate the probability of the union of $A$ and $B$ . Recall that the union of two events $A$ and $B$ is the set of outcomes that are classified as either $A$ OR $B$ .

At first glance, it might seem possible to calculate the probability of the union by simply adding the probability of $A$ to the probability of $B$ . This runs into trouble, however, whenever $A$ and $B$ have overlapping outcomes.

When adding $\mathbb{P}(A)$ and $\mathbb{P}(B)$ , the overlapping part of $A$ and $B$ (the intersection of $A$ and $B$ ) is counted twice. To compensate for this, subtract $\mathbb{P}(A \cap B)$ :

If two events $A$ and $B$ have overlapping outcomes, the probability of the union of $A$ and $B$ is calculated as follows:

$\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) − \mathbb{P}(A \cap B)$

If two events $A$ and $B$ are mutually exclusive, that is $A \cap B = \emptyset$ , the rule simplifies to:

$\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B)$

Consider the random experiment of rolling a single die. For this experiment, events $A$ and $B$ are defined as follows:

$A =$ 'you roll a number $\geq 4$ '
$B =$ 'you roll an even number'

What is $\mathbb{P}(A \cup B)$ ?

$\mathbb{P}(A \cup B) = \cfrac{4}{6}$

The probabilities of events $A$ and $B$ are:

$\mathbb{P}(A)=\cfrac{\text{'number of outcomes}\geq 4\text{'}}{\text{'total number of outcomes'}}=\cfrac{3}{6}$
$\mathbb{P}(B) = \cfrac{\text{'number of even outcomes'}}{\text{'total number of outcomes'}}=\cfrac{3}{6}$

In order to calculate the probability of the union, first calculate the probability of the intersection $\mathbb{P}(A \cap B)$ .

To determine how the intersection should be calculated, investigate whether events $A$ and $B$ are independent or not. If it is known that the outcome of the roll is a number $\geq 4$ , then the probability of the roll being even is $2$ out of $3$ ; namely $4$ and $6$ , but not $5$ :

$\mathbb{P}(B|A) =\cfrac{2}{3}$

This demonstrates that $\mathbb{P}(B) \neq \mathbb{P}(B|A)$ , so $A$ and $B$ are not independent. Therefore, the multiplication rule should be applied:

$\mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B|A) =\cfrac{3}{6}\cdot \cfrac{2}{3}=\cfrac{2}{6}$

Now, all the information needed for the calculation of the probability of the union of $A$ and $B$ is available:

$\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B) − \mathbb{P}(A \cap B) = \cfrac{3}{6}+\cfrac{3}{6}-\cfrac{2}{6}=\cfrac{4}{6}$

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