UK Secondary (7-11)
Mutually Exclusive and Non-Mutually Exclusive Events
Lesson

## 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) =1752152

=$\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:

Mutually exclusive events

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

Summary of probability relationships

P(A)+P(A') = 1

P(A U B) = P(A) + P(B) - P(AB)                 (non-mutually exclusive)

P(A U B) = P(A) + P(B)                  (mutually exclusive as P(AB)=0)

Here are some worked examples.

##### Question 1

A random card is picked from a standard deck. Find the probability that the card is:

1. red or a diamond

2. an ace or a diamond.

3. an ace of spades or an ace of clubs

4. a black or a face card

##### Question 2

Two events $A$A and $B$B are mutually exclusive.

If P(A) = $0.37$0.37 and P(A or B) = $0.73$0.73, what is P(B)?