Mutually exclusive events

Two events are said to be mutually exclusive if they cannot occur together; i.e., they have no elements in common. Non-mutually exclusive events can occur together; i.e., they have common elements.

For non-mutually exclusive events, the formula is:

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

where:

• $P(A \cup B)$ = Probability of event $A$ or event $B$ occurring
• $P(A \cap B)$ = Probability of both events $A$ and $B$ occurring

For mutually exclusive events, $P(A \cap B) = 0$ since the events cannot occur together, and the formula simplifies to:

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

Example 1

One card is selected from a pack of 52 playing cards. What is the probability that the card selected will be either a diamond or a black ace?

Solution:

$P(\text{diamond}) = \frac{13}{52}$

$P(\text{black ace}) = \frac{2}{52}$

$P(\text{diamond or black ace}) = P(\text{diamond}) + P(\text{black ace})$

$= \frac{13}{52} + \frac{2}{52} = \frac{15}{52}$

The probability is $\frac{15}{52}$.

Example 2

At a certain school, there are 30 teachers: 10 Physics, 8 Chemistry, and 5 Mathematics. If a teacher is selected at random, find the probability of getting:

a) a Physics teacher or a Chemistry teacher. b) a Chemistry teacher or a Mathematics teacher.

Solution:

Let $A$ = Physics, $B$ = Chemistry, $C$ = Mathematics.

$P(A) = \frac{10}{30} = \frac{1}{3}$

$P(B) = \frac{8}{30} = \frac{4}{15}$

$P(C) = \frac{5}{30} = \frac{1}{6}$

a)

$P(A \cup B) = P(A) + P(B) = \frac{1}{3} + \frac{4}{15} = \frac{5}{15} + \frac{4}{15} = \frac{9}{15} = \frac{3}{5}$

The probability of selecting a Physics or Chemistry teacher is $\frac{3}{5}$.

b)

$P(B \cup C) = P(B) + P(C) = \frac{4}{15} + \frac{1}{6} = \frac{8}{30} + \frac{5}{30} = \frac{13}{30}$

The probability of selecting a Chemistry or Mathematics teacher is $\frac{13}{30}$.

Example 3

$A$ and $B$ are mutually exclusive events such that $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{4}$. Find

a) $P(A \cup B)$ b) $P(A \cap B)$ c) $P(A' \cap B')$

Solution:

a)

$P(A \cup B) = P(A) + P(B) = \frac{1}{2} + \frac{1}{4} = \frac{2}{4} + \frac{1}{4} = \frac{3}{4}$

b) Since $A$ and $B$ are mutually exclusive, $P(A \cap B) = 0$.

c)

$P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) = 1 - \frac{3}{4} = \frac{1}{4}$