Mutually Exclusive and Non-Mutually Exclusive Events

Mutually exclusive events

If events are mutually exclusive, it means they cannot happen at the same time. 

Some examples of experiments that involve mutually exclusive events are:

• tossing a coin - Consider the events 'flipping a head' and 'flipping a tail'. You cannot flip a head and a tail at the same time. 
• rolling a die - Consider the events 'Rolling an even number' and 'rolling an odd number'. We can't roll any number which is both even and odd. 
• picking a card from a deck of cards - Consider the events 'Drawing a 7 card' and 'Drawing a 10 card'. They have no outcomes in common. There is no card that is both a 7 and a 10.

Since these events cannot both occur at the same time, they are mutually exclusive events.

 

However some events can happen at the same time and we call this non-mutually exclusive. For example:

• picking a card from a deck of cards - Consider the events 'drawing a Club card' and 'drawing a 7'. They have outcomes in common. We could pick a card that is a Club and a 7, because I could get the 7 of clubs. 
• rolling a die - Consider the events 'Rolling an even number' and 'Rolling a prime number'. They have outcomes in common, namely the number 2. 

Since these events can both occur at the same time,  they are non mutually exclusive events.

 

The mutually exclusive case

Consider a card experiment and the events A:'Drawing a 7 card' and B:'Drawing a 10 card'. What is P(A or B)?

We know that $P(event)=\frac{\text{number of favourable outcomes}}{\text{total possible outcomes}}$P(event)=number of favourable outcomestotal possible outcomes​.

Number of favourable outcomes = number of '7' cards + number of  '10' cards

                                                            = 4+4

                                                            = 8

Any double counting of favourable cards? No, because there are no cards that are both a '7' and a '10'.

So $\text{P(A or B) }=\frac{8}{52}$P(A or B) =852​

                     = $\frac{2}{13}$213​

Note: $\text{P(A) }+\text{P(B) }=\frac{4}{52}+\frac{4}{52}$P(A) +P(B) =452​+452​

                               =$\frac{8}{52}$852​

                               =$\frac{2}{13}$213​

 

In this mutually exclusive case: P(A or B)=P(A)+P(B)

 

The non mutually exclusive case

Consider a card experiment and the events A:'Drawing a Club card' and B:'Drawing a 7 card'. What is P(A or B)?

We know that $\text{P(event) }=\frac{\text{number of favourable outcomes}}{\text{total possible outcomes}}$P(event) =number of favourable outcomestotal possible outcomes​.

Number of favourable outcomes = number of 'club' cards + number of  '7' cards

                                                            = 13+4

                                                            = 17

Any double counting of favourable cards? Yes, because there is 1 card which is the 7 of clubs. We have counted it twice - once as a 'club' card and once as a '7' card. So there are actually only 16 favourable outcomes.

We actually came upon this idea when we looked at Venn Diagrams, and we can see the same applies here.

 

So,

$\text{P(A or B) }$P(A or B) =  $\frac{16}{52}$1652​

                 = $\frac{4}{13}$413​

 

Note:

•  $\text{P(A) }+\text{P(B) }=\frac{13}{52}+\frac{4}{52}$P(A) +P(B) =1352​+452​ 

                                 = $\frac{17}{52}$1752​

•  $\text{P(A and B) }=\frac{1}{52}$P(A and B) =152​    (There is 1 card which is a club AND a 7)
•  $\text{P(A) }+\text{P(B) }-\text{P(A and B) }=\frac{17}{52}-\frac{1}{52}$P(A) +P(B) −P(A and B) =1752​−152​

                                                        =$\frac{16}{52}$1652​                                                                                          

                                                        =$\frac{4}{13}$413​

 

In this non mutually exclusive case: P(A or B)=P(A)+P(B)-P(A and B)

 

Probability relationships

Remember our probability relationship:

P(A U B) = P(A) + P(B) - P(A and B)

As mutually exclusive events CANNOT happen together. We can say that P(A and B) = 0.

So this means that:

 

 

Here are some worked examples.

Question 1

Question 2