3. Probability: Probability
Probability of the Difference
The probability of the intersection is not only used to calculate the probability of the union, but it is also used to calculate the probability of the difference. Recall that the difference of events #A# and #B# are all outcomes classified as #A#, but NOT as #B#: #A\backslash B#.
The difference of #A# and #B# thus contains all outcomes that are classified as #A#, minus the outcomes in #A# that are also classified as #B#.
So the difference of #A# and #B# is #A# minus the intersection of #A# and #B#:
\[\mathbb{P}(A\backslash B) = \mathbb{P}(A) − \mathbb{P}(A \cap B)\]
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 #\leq 3#'
- #B =# 'you roll an odd number'
#\mathbb{P}(A \backslash B) = \cfrac{1}{6}#
The probabilities of events #A# and #B# are:
- #\mathbb{P}(A)=\cfrac{\text{'number of outcomes}\leq 3\text{'}}{\text{'total number of outcomes'}}=\cfrac{3}{6}#
- #\mathbb{P}(B) = \cfrac{\text{'number of odd outcomes'}}{\text{'total number of outcomes'}}=\cfrac{3}{6}#
In order to calculate the probability of the difference, 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 #\leq 3#, then the probability of the roll being odd is #2# out of #3#; namely #1# and #3#, but not #2#:
\[\mathbb{P}(B|A) =\cfrac{2}{3}\]
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, the probability of the difference of #A# and #B# can be calculated:
\[\mathbb{P}(A \backslash B) = \mathbb{P}(A) - \mathbb{P}(A \cap B) = \cfrac{3}{6}-\cfrac{2}{6}=\cfrac{1}{6}\]
