3. Probability: Probability
Probability of the Intersection
Quite often, it is needed to calculate the probability of the intersection of two events #A# and #B#. Recall that the intersection #A# and #B# is the set of outcomes that are classified as #A# AND #B#.
To calculate the probability of the intersection of two events, apply the following rules:
Rules
- Multiplication rule: The probability that events #A# and #B# both occur is equal to the probability that event #A# occurs multiplied by the probability that event #B# occurs, given that #A# has occurred:
#\mathbb{P}(A\cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B|A)# - If #A# and #B# are independent, then
#\mathbb{P}(A\cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B)#
Consider the random experiment of rolling two dice. Let #A# be the event of 'getting at least one 6'.
What is #\mathbb{P}(A)#?
#\mathbb{P}(A) =\dfrac{11}{36}#
One way to calculate the probability of 'getting at least one 6' is to make use of the complement rule:
\[\mathbb{P}(A) = 1 − \mathbb{P}(A^c)\]
Here, #A^c# is the event of getting no sixes in two rolls.
To calculate the value of #\mathbb{P}(A^c)#, define events #B_1# and #B_2# as follows:
- #B_1=# 'getting no 6 on the first roll'
- #B_2=# 'getting no 6 on the second roll'
The probabilities corresponding to these events are:
- #\mathbb{P}(B_1) = \cfrac{5}{6}#
- #\mathbb{P}(B_2) = \cfrac{5}{6}#
The event of getting no sixes on two rolls equals the event of getting no six on the first roll AND getting no six on the second roll:
\[A^c = B_1 \cap B_2\]
Since dice rolls are independent events, rule 2 can be applied to calculate the probability of the intersection of #B_1# and #B_2#:
\[\mathbb{P}(A^c)=\mathbb{P}(B_1 \cap B_2) = \mathbb{P}(B_1) \cdot \mathbb{P}(B_2) = \frac{5}{6}\cdot \frac{5}{6}=\frac{25}{36}\]
Now that #\mathbb{P}(A^c)# is known, #\mathbb{P}(A)# can be calculated:
\[\mathbb{P}(A) = 1 − \mathbb{P}(A^c) =\frac{11}{36}\]
