Addition Rule

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) = $\frac{1}{26}$ + $\frac{1}{26}$ = $\frac{2}{26}$ = $\frac{1}{13}$.