Probability of the Intersection

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:

$\mathbb{P}(A) = 1 − \mathbb{P}(A^c) =\frac{11}{36}$

New example