Probability Addition Rule - Formula & Examples

Probability is an idea which helps to find how certain events are to happen. It is based on observations of certain events. Probability is one of the most important topics of the mathematics. Probability of an certain event, say $A$ is symbolized by $P(A)$. Always it lies between  0 and 1.

We define probability as the number of outcomes of an event associated with a random experiment divided by the total number of outcomes possible in that experiment. Let $n (S)$ denote the number of outcomes of the experiment and A be an event whose outcomes are $n (A)$. Then, probability of occurrence of event A is given by: $P (A)$ = $\frac{n (A)}{n (S)}$.

Suppose we have two events X and Y associated with a random experiment E. then the probability of occurrence of either X or Y or both is given by:

$P (X\ or\ Y)$ = $P (X) + P (Y)$ – $P (X\ and Y)$

$\rightarrow$ $P (X \cup Y)$ = $P (X) + P (Y)$ - $P (X \cap Y)$

Where $P (X)$ is the probability of occurrence of event $X, P (Y)$ is the probability of occurrence of event $Y$ and $P (X\ and\ Y)$ [also written as $P (X\ \cap Y)$ is the probability of occurrence of both $X$ and $Y$ events simultaneously.

When the two events $X$ and $Y$ are mutually exclusive that is the occurrence of event $X$ does not affects the probability occurrence of event Y or vice versa. In this case the intersection probability that is $P (X \cap Y)$ = 0. And we have,

$P (X\ or\ Y)$ = $P (X)$ + $P (Y)$

## Word Problems

Some examples on probability addition rule are illustrated below:

Example 1:
We have rolled a six sided die. Find the probability of getting an even number or a 5.

Solution:

Let E be the experiment of rolling a die and S be the sample space, then

S = {2, 4, 1, 5, 3, 6} => n (S) = 6

A = event of getting an even number = {4, 2, 6} => n (A) = 3

B = event of getting a 5 = {5} => n (B) = 1

So we have P (A) = $\frac{3}{6}$, P (B) = $\frac{1}{6}$.

It is clear that the two events are mutually exclusive, hence $P (A \cap B)$ = 0

So according to the probability addition rule we have:

$P (A\ or\ B) =$\frac{3}{6}$+$\frac{1}{6}$=$\frac{4}{6}$=$\frac{2}{3}$Example 2: We are given a deck of 52 cards. We pick a card from it. Find the probability of getting a heart or a red king. Solution: Let S be the sample space then$n (S)$= 52.$A$= event of getting a red king =>$n (A)$= 2 =>$P (A)$=$\frac{2}{52}B$= event of getting a heart =>$n (B)$= 13 =>$P (B)$=$\frac{13}{52}$Clearly$A$and$B$are not mutually exclusive as a red king can also be of heart. =>$n (A\  \cap\  B)$= 1 =>$P (A \cap B)$=$\frac{1}{52}$. By addition rule of probability we have: P (A or B) =$\frac{2}{52}$+$\frac{13}{52}\frac{1}{52}$=$\frac{14}{52}$=$\frac{7}{26}$Example 3: What is the probability of getting a total of 3 or 7, when two dice are rolled? Solution: Let event A be the total of 7 and event B the total of 3. When we roll two dice we have 36 events in a sample space. Number of events getting total of 3 = 2 Number of events getting total of 7 = 6 =>$P(A)$=$\frac{6}{36}$and$P(B)$=$\frac{2}{36}$Both the events are mutually exclusive since they do not have any common event. i.e.$P(A \cap B)$= 0 Now$P(A \cup B)$=$P(A)$+$P(B)$=$\frac{6}{36}$+$\frac{2}{36}$=$\frac{8}{36}$=$\frac{2}{9}\$.