### Chapter 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 die with six sides, numbered from #1# to #6#, and observing the number on top. For this experiment, we define the following events:

- #A =# 'The number is greater than or equal to #4#'
- #B =# 'The number is even'

Calculate the probability that both of these events will occur.

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

- #\mathbb{P}(A)=\cfrac{\text{number of outcomes}\geq 4}{\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, we first need to calculate the probability of the intersection #\mathbb{P}(A \cap B)#. To do this, we need to know whether events #A# and #B# are independent.

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 we must conclude #A# and #B# are not independent. As such, the probability of the intersection is calculated as follows:

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

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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