If events are mutually exclusive, it means they cannot happen at the same time.
Some examples of experiments that involve mutually exclusive events are:
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:
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(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)
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:
= $\frac{17}{52}$1752
=$\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:
P(A U B) = P(A) + P(B)
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)?