Probability is defined as the total number of outcomes of an event that is associated with an experiment divided by the total number of outcomes that are possible as a result of that experiment. “n (S)” denotes the number of outcomes of the experiment where ‘S’ is the sample space that is a set of all possible outcomes and A an event whose number of outcomes is given by n (A).

Then, the probability of occurrence of the event A, P (A) = $\frac{n (A)}{n (S)}$.

## Formula

Consider A and B, two events, associated with an experiment E. Then the probability of occurrence of either A or B or occurrence of both is given by:

P (A or B) = P (A) + P (B) - P (A and B)

Or

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

Where P (A) is the probability of event A, P (B) is the probability of event B and P (A and B) [or P $(A \cap B)$] is the probability of both A and B occurring simultaneously.

**Now two cases arise:**

First Case:

If A and B are not mutually exclusive, that is, the occurrence of event A does affect the probability of occurrence of the event B or vice versa, then the probability of intersection, that is, P (A ⋂ B) ≠ 0, this implies,

P (A or B) = P (A) + P (B) – P (A $\cap$ B)

Second Case:

If A and B are mutually exclusive events, that is, the occurrence of event A does not affect the probability of occurrence of the event B and vice versa, then the probability of intersection, that is, P (A $\cap$ B) = 0, this implies,

P (A or B) = P (A) + P (B)

## Problems

**Let us see some examples on addition rule of probability:**

**Example 1:** A six sided fair die is rolled. Find the probability of getting an odd number or a prime number.

**Solution:**

E: rolling a die experiment

S: sample space

$\rightarrow$ S = {6, 4, 1, 3, 5, 2}

$\rightarrow$ n (S) = 6

M = event of getting an odd number = {1, 5, 3}

$\rightarrow$ n (M) = 3

N = event of getting a prime number = {2, 3, 5}

$\rightarrow$ n (N) = 3

M $\cap$ N = {3, 5}

$\rightarrow$ n (M $\cap$ N) = 2

So we have P (M) = $\frac {3}{6}$ =$\frac{1}{2}$, P (N) = $\frac{3}{6}$ = $\frac{1}{2}$ and P (M$\cap$ N) = $\frac{2}{6}$ = $\frac{1}{3}$.

So by the probability addition rule we get:

P (A or B) = $\frac{1}{2}$ + $\frac{1}{2}$ - $\frac{1}{3}$ = 1 - $\frac{1}{3}$ = $\frac{2}{3}$.

**Example 2:** We pick a card from a deck of 52 cards. Find the probability of getting a red ace or a black king.

**Solution:**

Clearly, n (S) = 52.

A = event of getting a black king

$\rightarrow$ n (A) = 2

$\rightarrow$ P (A) = $\frac{2}{52}$ = $\frac{1}{26}$

B = event of getting a red ace

$\rightarrow$ n (B) = 2

$\rightarrow$ P (B) = $\frac{2}{52}$ = $\frac{1}{26}$

Clearly A and B are mutually exclusive events. Hence by the addition rule of probability we have:

P (A or B) = $\frac{1}{26}$ + $\frac{1}{26}$ = $\frac{2}{26}$ = $\frac{1}{13}$.