In this post we will discuss the concept of mutually non – exclusive events in probability.

The concept is extensively used in the probability chapter so make sure you understand the concept well. At the end, solved problems are also provided for conceptual clarity.

## Definition of mutual non exclusive events

Two events are said to be mutually non exclusive when **there is one or more common element between the two**.

In case of non exclusive events, **both event can occur simultaneously in the experiment**.

**For example**, Consider the experiment of throwing a dice.

Let A be the event of getting even number.

A = {2, 4, 6 }

Let B be the event of getting number less than 3.

B = { 1, 2 }

On comparing event A & B, note that there is common element “2” present in both the sets. Hence, both the events are not mutually exclusive.

### Formula for mutually non exclusive events

If A & B are mutually non exclusive events then the **probability of A union B **is given by following formula;

\mathtt{P\ ( A\ \cup \ B\ ) \ =\ P( A) \ +\ P( B) -P( A\cap B)}\\\ \\

The probability of A union B is the sum of individual probabilities minus probability of A intersection B.

You have to memorize the formula for solving questions related to the concept.

### Questions on Mutually non exclusive events

**Question 01**

In a box, there are 15 balls with number 1 to 15 imprinted on them such that each number is non repeatable. A ball is randomly drawn from the box. Find the probability that the number is multiple of 2 or 3.

**Solution**

Since number 1 to 15 are present in the box, the sample space is given as;

S = {1, 2, 3, 4, 5, 6, 7, 8 , 9, 10, 11, 12, 13, 14, 15}

**Let A be the event of getting multiple of 2.**

A = {2, 4, 6, 8, 10, 12, 14 }

Total number of elements in A = 7

Probability P(A) = 7 / 15

**Let B be the event of getting multiple of 3**.

B = {3, 6, 9, 12, 15}

Total number of elements in B = 5

Probability P (B) = 5 / 15

**Note that element 6 and 12 are common element in set A & B.**

Total number of common elements = 2

P (A ∩ B) = 2/ 15

Now applying mutually non exclusive formula to calculate the probability of getting multiple of 2 or 3.

\mathtt{P\ ( A\ \cup \ B\ ) \ =\ P( A) \ +\ P( B) -P( A\cap B)}\\\ \\ \mathtt{P\ ( A\ \cup \ B\ ) \ =\ \frac{7}{15} +\frac{5}{15} -\frac{2}{15}}\\\ \\ \mathtt{P\ ( A\ \cup \ B\ ) \ =\ \frac{10}{15} =\frac{2}{3}}

Hence, **2/3 is the required probability.**

**Question 02**

In an experiment, two dice are thrown and numbers are recorded. Consider the following events;

A = sum of number is greater than 9

B = number 5 on either dice

C = sum is multiple of 7

Check which of the given pair is mutually non exclusive.**Solution**

Let’s first write the sample space for the experiment.

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

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6) }

Now let’s write the set form of given events;**A = sum of number is greater than 9**

A = { (4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6) }

**B = number 5 on either dice **

B = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (1, 5), (2, 5), (3, 5), (4, 5), (6, 5)

**C = sum is multiple of 7 **

C = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

**Comparing events A & B**

Note that (5, 5), (5, 6) and (6, 5) are the common elements present in both events. Hence, both events are mutually non exclusive.

**Comparing events B & C**

Note that (2, 5) is common element present in B & C. Hence, both are mutually non exclusive events.**Comparing events C & A**

There is no common element between C & A. Hence, they are mutually exclusive.

**Question 03**

Consider the experiment of throwing a dice. Find the probability of getting odd number or number greater than 3.

**Solution**

Let’s first write the sample space for the experiment.

S = {1, 2, 3, 4, 5, 6 }

Let **A be the event of getting odd number** and **B be the event of getting number greater than 3**.

A = { 1, 3, 5 }**P(A) = 3/6**

B = { 4, 5, 6 }**P(B) = 3/6**

Note that both events A & B are mutually non exclusive events. The number 5 is common in both set A & B.**P ( A ∩ B ) = 1/6**

Now using the mutually non exclusive formula;

\mathtt{P\ ( A\ \cup \ B\ ) \ =\ P( A) \ +\ P( B) -P( A\cap B)}\\\ \\ \mathtt{P\ ( A\ \cup \ B\ ) \ =\ \frac{3}{6} +\frac{3}{6} -\frac{1}{6}}\\\ \\ \mathtt{P\ ( A\ \cup \ B\ ) \ =\ \frac{5}{6}} \\\ \\

Hence, **5/6 is the required probability.**

**Question 04**

From the deck of 52 cards, one card is selected at random. What is the probability of getting spade or a Jack.**Solution**

Let** A is the event of getting a spade** and **B be the event of getting a Jack**.

A = { ” 13 cards of spade ” }**P(A) = 13/52**

B = {” 4 cards of Jack ” }**P(B) = 4/52**

We know that there is 1 ” Jack of Spade” in the pack of 52 cards. Hence, there is one common element between event A & B.**P (A ∩ B) = 1 / 52**

Now using the formula for mutually non exclusive event.

\mathtt{P\ ( A\ \cup \ B\ ) \ =\ P( A) \ +\ P( B) -P( A\cap B)}\\\ \\ \mathtt{P\ ( A\ \cup \ B\ ) \ =\ \frac{13}{52} +\frac{4}{52} -\frac{1}{52}}\\\ \\ \mathtt{P\ ( A\ \cup \ B\ ) \ =\ \frac{16}{52}} \\\ \\

Hence, **16/52 is the required probability**.