Addition of Probabilities

Probability → Vocabulary → Formula → Example Questions → Addition of Probabilities

First consider what are joint (or not-mutually exclusive) and disjoint (or mutually exclusive) events.

Suppose, we have sample space $S.S=\{1,2,3,4,5,6\}$

Then even and odd numbers in the above set are disjoint events. As Even Numbers = 2, 4,6 and Odd Numbers = 1, 3, 5

But, if we are interested in even numbers and multiples of three, then
Even Numbers = 2, 4, 6 and
Multiples of Three = 3, 6
So, 6 is a joint element.

Addition Rule: Disjoint Events

$P(A\cup B)=P(A)+P(B)$

Addition Rule: Joint Events

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Remember:

$P(A \textrm{ or } B)=P(A\cup B)$

Example 1. A dice is rolled once. What is the probability that the number shown is 3 or even number?
Solution
3 is one event out of total 6 events, and even numbers {2,4,6} are three events out of total six events. Thus their probabilities are

$P(3)=\frac{1}{6}=0.1667$ , and

$P(\textrm{even})=\frac{3}{6}=0.5$

Thus,

$P(3 \textrm{ or even})=P(3)+P(\textrm{even})=0.1667+0.5=0.667$

Example 2. A dice is rolled once. What is the probability that the number shown is 3 or odd number?
Solution

$P(3)=\frac{1}{6}=0.1667$ , and

$P(\textrm{odd})=\frac{3}{6}=0.5$

Now, there is one joint element i.e. 6. So,

$P(3\cap \textrm{odd})=\frac{1}{6}=0.1667$

Thus,

$P(3 \textrm{ or odd})=P(3)+P(\textrm{odd})-P(3\cap \textrm{odd})=0.1667+0.5-0.1667=0.5$