Mutually Inclusive Events Examples

# Mutually Inclusive Events Examples

Mutually Inclusive Events are those events which are common outcomes in a set of given events. The best way to explain mutually inclusive events is by taking up an example one event being even numbers we get on throwing a dice and the second event being numbers less than $5$ on throwing a dice. If the first event is called '$A$' and the second event is called '$B$', then $A$ = $\{ 2, 4, 6 \}$ and $B$ = $\{ 1, 2, 3, 4 \}$. Outcome which are common to both the events are $A \cap B$ = $\{ 2, 4 \}$ and $P(A \cap B)$ = $\frac{2}{6}$ = $\frac{1}{3}$. We can even draw Venn diagram to represent the example properly

Formula which can be applied in case of mutually inclusive event is

$P(A\ or\ B)$ = $P(A) + P(B) â€“ P(A\ and\ B)$

$P(A \cup B)$ = $P(A) + P(B) â€“ P(A \cap B)$

Where, $P(A \cup B)$ is the combined probability of occurrence of either $A$ or $B,\ P(A)$ is the probability of occurrence of event $A,\ P(B)$ is the probability of occurrence of event $B$ and $P(A \cap B)$ is the probability of joint occurrence of event $A$ and event $B$

## Examples

Example 1:

A fair dice is tossed. What is the probability of tossing an even number or a number greater than $3$?

Solution:

Let event $A$ be getting an even number. So $A$ = $\{ 2, 4, 6 \}$, $n(A)$ = $3$ and event $B$ be getting numbers greater than $3$. So, $B$ = $\{ 4, 5, 6 \}$, $n(B)$ = $3$. The number of outcomes in the sample space is $n(S)$ = $6$. The probability of event $A$ would be $P(A)$ =  $\frac{n(A)}{n(S)}$ = $\frac{3}{6}$ = $\frac{1}{2}$. Similarly, the probability of event $B$ would be $P(B)$ = $\frac{n(B)}{n(S)}$ = $\frac{3}{6}$ = $\frac{1}{2}$. The common outcome from both the events would be the number $4$, so the probability of joint occurrence of event $A$ and event $B$ is $P(A \cap B)$ = $\frac{1}{6}$. Thus, the probability of tossing an even number or a number greater than $3$ is

$P(A \cup B)$ = $P(A) + P(B) â€“ P(A \cap B)$

$P(A \cup B)$ = $\frac{1}{2}$ $+$ $\frac{1}{2}$ $â€“$ $\frac{1}{6}$

$P(A \cup B)$ = $\frac{5}{6}$
Example 2:

What is the probability of choosing a card from a deck of cards that is a heart or a nine?

Solution:

Let event $A$ be choosing a heart from the deck of cards. As there are $13$ heart cards in a deck of cards, so the probability of selecting a heart card would be $P(A)$ = $\frac{n(A)}{n(S)}$ = $\frac{13}{52}$ = $\frac{1}{4}$. Let event $B$ be choosing a nine from a deck of cards. As there are $4$ nineâ€™s in a deck of cards, so the probability of choosing a nine card shall be $P(B)$ = $\frac{n(B)}{n(S)}$ = $\frac{4}{52}$ = $\frac{1}{13}$. The total number of cards present in a deck of cards is $52$, so the number of elements in the sample space would be $n(S)$ = $52$. The common outcome from both the events that is selecting a heart and that too nine in number is1, so the probability of joint occurrence of event $A$ and event $B$ is $P(A \cap B)$ = $\frac{1}{52}$. Thus, the probability of choosing a card from a deck of cards that is a heart or a nine is

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

$P(A\ or\ B)$ = $\frac{1}{4}$ $+$ $\frac{1}{13}$ $-$ $\frac{1}{52}$

$P(A\ or\ B)$ = $\frac{16}{52}$

$P(A\ or\ B)$ = $\frac{4}{13}$
Example 3:

What is the probability of choosing a number from $1$ to $15$ that is less than $10$ or even?

Solution:

Let event $A$ be choosing a number from 1 o 15 that is less than $10$. So, $A$ = $\{ 1, 2, 3, 4, 5, 6, 7, 8, 9 \}$ and $n(A)$ = $9$. Let event $B$ be choosing an even number from $1$ to $15$. So, $B$ = $\{ 2, 4, 6, 8, 10, 12, 14 \}$ and $n(B)$ = $7$. The number of outcomes in the sample space are the numbers from $1$ to $15$, $n(S)$ = $15$. The probability of event $A$ would be $P(A)$ = $\frac{n(A)}{n(S)}$ = $\frac{9}{15}$. Similarly, the probability of event $B$ would be $P(B)$ = $\frac{n(B)}{n(S)}$ = $\frac{7}{15}$. The common outcome from both the events would be the number $A \cap B$ = $\{ 2, 4, 6, 8 \}$ , so the probability of joint occurrence of event $A$ and event $B$ is $P(A \cap B)$ = $\frac{4}{15}$. Thus, the probability of choosing a number from $1$ to $15$ that is less than $10$ or even is

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

$P(A\ or\ B)$ = $\frac{9}{15}$ $+$ $\frac{7}{15}$ $-$ $\frac{4}{15}$

$P(A\ or\ B)$ = $\frac{12}{15}$

$P(A\ or\ B)$ = $\frac{4}{5}$
Example 4:

$2$ fair dice are rolled. What is the probability of getting a sum less than $8$ or a sum equal to $9$?

Solution:

Let us first draw the sum table when two fair dice are rolled

Let event $A$ be getting a sum less than $8$ on rolling two fair dice. In the table above those numbers are shaded in brick red color, so $n(A)$ = $21$. Let event $B$ be getting a sum equal to $9$ on rolling two fair dice. In the table above those numbers are shaded in green color, so $n(B)$ = $4$. The number of outcomes in the sample space is the total sum of numbers we get on rolling two fair dice, $n(S)$ = $36$. The probability of event $A$ would be $P(A)$ = $\frac{n(A)}{n(S)}$ = $\frac{21}{36}$. Similarly, the probability of event $B$ would be $P(B)$ = $\frac{n(B)}{n(S)}$ = $\frac{4}{36}$. As there is no elements in common so $A \cap B$ = $0$. In this type of condition the events are said to be mutually exclusive events. Thus, the probability of getting a sum less than $8$ or a sum equal to $9$

$P(A \cup B)$ = $P(A) + P(B) â€“ P(A \cap B)$

$P(A \cup B)$ = $\frac{21}{36}$ $+$ $\frac{4}{36}$ $-$ $0$

$P(A \cup B)$ = $\frac{25}{36}$
Example 5:

What is the probability of getting a sum less than $5$ and a sum less than $4$ when two fair dice is rolled?

Solution:

Let us first draw the sum table when two fair dice are rolled.

Let event $A$ be getting a sum less than $5$ on rolling two fair dice. In the table above those numbers are shaded in purple color, so $n(A)$ = $6$. Let event $B$ be getting a sum less than $4$ on rolling two fair dice. In the table above those numbers are shaded in pink color, so $n(B)$ = $3$. The number of outcomes in the sample space is the total sum of numbers we get on rolling two fair dice, $n(S)$ = $36$. The probability of event $A$ would be $P(A)$ = $\frac{n(A)}{n(S)}$ = $\frac{6}{36}$. Similarly, the probability of event $B$ would be $P(B)$ = $\frac{n(B)}{n(S)}$ = $\frac{3}{36}$. The common outcome from both the events would be the number $A \cap B$ = $\{ 2, 3, 3 \}$, so the probability of joint occurrence of event $A$ and event $B$ is $P(A \cap B)$ = $\frac{3}{36}$. Thus, the probability of getting a sum less than $5$ or a sum less than $4$.

$P(A\ or\ B)$ = $P(A) + P(B) - P(A and B)$P(A\ or\ B)$=$\frac{6}{36}+\frac{3}{36}-\frac{3}{36}P(A\ or\ B)$=$\frac{6}{36}P(A\ or\ B)$=$\frac{1}{6}\$