intersection of events represents outcomes that occur in all specified events simultaneously. For events $A$ and $B$, the intersection is denoted $A\cap B$.

Definition: $A\cap B$ includes outcomes that are common to both $A$ and $B$.

Probability: The probability of the intersection depends on whether the events are independent or dependent:

• Independent Events: If $A$ and $B$ are independent (the occurrence of one does not affect the other), then:

$$P(A\cap B)=P\left(A\right)·P\left(B\right)$$

• Dependent Events: If not independent, use conditional probability:

$$P(A\cap B)=P\left(A\right)·P\left(B\right|A)=P(B)·P(A\left|B\right)$$

Question:

Two dice are rolled. What is the probability that the first die shows an even number and the second die shows a 5?

Solution:

Let:

$P\left(\text{even on first die}\right)=\frac{3}{6}=\frac{1}{2}$

$P\left(\text{5 on second die}\right)=\frac{1}{6}$

Since rolls are independent (we will study about independent events in next article):

$$P\left(\text{even and 5}\right)=\frac{1}{2}·\frac{1}{6}=\frac{1}{12}$$

There is a $\frac{1}{12}$ or 8.33% chance both conditions are met.

The union of two or more events represents the occurrence of at least one of the events. For two events $A$ and $B$, the union is denoted $A\cup B$.

Definition: $A\cup B$ includes all outcomes in the sample space that belong to $A$, $B$, or both.

Probability: The probability of the union is given by the addition rule:

$$P(A\cup B)=P\left(A\right)+P\left(B\right)–P(A\cap B)$$

The term $P(A\cap B)$ is subtracted to avoid double-counting outcomes that belong to both $A$ and $B$.

Question:

In a class of 20 students, 12 are girls, and 5 students wear glasses. If 3 of the girls wear glasses, what is the probability that a randomly selected student is either a girl or wears glasses?

Solution:

Let:

$P\left(\text{girl}\right)=\frac{12}{20}$

$P\left(\text{glasses}\right)=\frac{5}{20}$

$P\left(\text{girl and glasses}\right)=\frac{3}{20}$

Apply the union formula:

$$P\left(\text{girl or glasses}\right)=P\left(\text{girl}\right)+P\left(\text{glasses}\right)–P\left(\text{girl and glasses}\right)=\frac{12}{20}+\frac{5}{20}–\frac{3}{20}=\frac{14}{20}=0.7$$

There is a 70% chance the student is either a girl or wears glasses

For Multiple Events: For three events $A$, $B$, and $C$, the formula extends to:

$$P(A\cup B\cup C)=P\left(A\right)+P\left(B\right)+P\left(C\right)–P(A\cap B)–P(B\cap C)–P(A\cap C)+P(A\cap B\cap C)$$

This is the inclusion-exclusion principle.

The complement of an event $A$, denoted $A^c$ or $\overline{A}$, represents all outcomes in the sample space that are not in $A$.

Definition: If $S$ is the sample space, then $A^c=S\backslash A$.

Probability: Since $A$ and $A^c$ cover the entire sample space and are mutually exclusive (we will discuss about mutually exclusive events in next article):

$$P\left(A\right)+P\left(A^c\right)=1⟹P\left(A^c\right)=1–P\left(A\right)$$

Question:

A day is chosen at random. What is the probability that the chosen day is not a weekday?

Solution:

There are 5 weekdays (Mon–Fri), so:

$P\left(\text{weekday}\right)=\frac{5}{7}$

$$P\left(\text{not weekday}\right)=1–\frac{5}{7}=\frac{2}{7}$$

The addition rule is used to compute the probability of the union of events, as introduced above. It adjusts for overlapping outcomes.

General Form: For two events:

$$P(A\cup B)=P\left(A\right)+P\left(B\right)–P(A\cap B)$$

Mutually Exclusive Events: If $A$ and $B$ cannot occur together ($A\cap B=\varnothing$), then:

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

Question:

A bag contains 3 red, 4 green, and 2 blue marbles. One marble is drawn. What is the probability it is either red or green?

Solution:

Red and green are mutually exclusive.

$P\left(\text{red}\right)=\frac{3}{9}$

$P\left(\text{green}\right)=\frac{4}{9}$

$$P\left(\text{red or green}\right)=\frac{3}{9}+\frac{4}{9}=\frac{7}{9}$$

The difference $A\backslash B$ represents outcomes in $A$ that are not in $B$.

Probability:

$$P(A\backslash B)=P\left(A\right)–P(A\cap B)$$

Example question:

In a class of 30 students, 10 play football. Among them, 4 also play cricket. What is the probability that a randomly selected student plays only football?

Solution:

$P\left(\text{football}\right)=\frac{10}{30}=\frac{1}{3}$

$P\left(\text{football and cricket}\right)=\frac{4}{30}$

So,

$$P\left(\text{only football}\right)=\frac{10}{30}–\frac{4}{30}=\frac{6}{30}=\frac{1}{5}$$

There is a 1 in 5 chance the student plays only football.