Probability

Hong Kong

Stage 4 - Stage 5

Lesson

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.**

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`(`e``v``e``n``t`)=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)**

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)**

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.

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

red or a diamond

an ace or a diamond.

an ace of spades or an ace of clubs

a black or a face card

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*)?