Our world is full of uncertainty and probable events; such as storms, accidents, noisy communications, financial markets etc. We come across with data everywhere. Statistical interference and probabilistic modeling are the fundamental ways of analyzing data and making scientific predictions. Probability is a theory of predicting the chances of certainty or uncertainty of the events. In probability theory, an event is defined as an outcome of some experiment. For example – obtaining a head or a tail on the toss of a coin. Broadly speaking, the events are of two types: mutually exclusive and non-mutually exclusive. Let us go ahead and learn about non-mutually exclusive events in detail. 


In probability, we come across with the events that cannot happen simultaneously and the events that can happen simultaneously. The first ones are known as mutually exclusive events, while the latter ones are non-mutually exclusive events.

The two events that can occur at the same time are called non-mutually exclusive events. For example – probability of getting an even number and a number greater than 3 on a roll of die are two non-mutually exclusive evens since the an even number on a dice can be greater than 3 as well.

Sample space for even numbers on a die = {2, 4, 6}

Sample space for numbers greater than 3 on a die = {4, 5, 6}

We can see that few elements in both the sample spaces are common. This indicates that these two are non-mutually exclusive events.

When two events have at least one common outcome in them, they are known as non mutually exclusive. Let us suppose that two events P and Q are non mutually exclusive, then it is certain that they are having some outcomes in common. So,

$P \cap Q \neq \phi$

Similarly, if P, Q and R be three non mutually exclusive events then

$P \cap Q \cap R \neq \phi$

Hence, there must be at least one common point.

The following Venn diagram indicates two non mutually exclusive events P and Q. 

non mutually exclusive event

The area which is common to both P and Q is indicating P $\cap$ Q which is not a null set.

The formula for finding probability of non mutually exclusive events Q and R is:

P(Q $\cup$ R) = P(Q) + P(R) -  P(Q $\cap$ R)

For three events non mutually exclusive events

P(Q $\cup$ R $\cup$ S) = P(Q) + P(R) + P(S) -  P(Q $\cap$ R) - P(R $\cap$ S) - P(S $\cap$ Q) - P(Q $\cap$ R $\cap$ S)

What is the Difference Between Mutually Exclusive and Non-Mutually Exclusive Events?

The main difference between both of them is that the mutually exclusive events do not occur at the same time and non mutually exclusive events do happen at the same time. First has no common element, while latter must have at least one common element. This is the reason the mutually exclusive events are also termed as disjoint events.

The following Venn diagrams clearly explain the difference between mutually exclusive and non mutually exclusive events.

non mutually

Example 1: Find the probability of obtaining an odd number or a prime number on a roll of die.

Solution: Total number of outcomes n(E) = 6

Let the event of obtaining an odd number be represented by A and obtaining a prime number be expressed by B.

Sample space for odd number = {1, 3, 5}

Probability of obtaining an odd number P(A) = $\frac{3}{6}$

Sample space for prime number = {2, 3, 5}

Probability of obtaining an odd number P(A) = $\frac{3}{6}$

(A $\cap$ B) = {3, 5}

P(A $\cap$ B) = $\frac{2}{6}$

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

= $\frac{3}{6}$ + $\frac{3}{6}$ - $\frac{2}{6}$

= $\frac{4}{6}$ = $\frac{2}{3}$

Example 2: A card is drawn at random from a well-shuffled pack of cards. What is the probability of getting either a red card or a face card?

Solution: Total number of cards n(E)= 52

Total number of red cards in a pack = n(R) = 26

[Note: there are 13 cards of 4 suits each and 2 suits are of red color, so 26 red cards]

Probability of getting a red card P(R) = $\frac{26}{52}$ 

Number of face cards n(F) = 12

[Face cards are jack, queen and king for all 4 suits, i.e 12 face cards]

Probability of getting a face card P(F) = $\frac{12}{52}$

n(R and F) = number of cards that are red as well as face cards = 6

[Note: 3 face card in 2 red suits]

P(R $\cap$ F) = = $\frac{6}{52}$

Now,

P(R $\cup$ F) = P(R) + P(F) – P(R $\cap$ F)

P(R $\cup$ F) = $\frac{26}{52}$ + $\frac{12}{52}$ – $\frac{6}{52}$ = $\frac{32}{52}$

= $\frac{8}{13}$