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