NZ Level 8 (NZC) Level 3 (NCEA) [In development]
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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)?

 

Outcomes

S8-4

Investigate situations that involve elements of chance: A calculating probabilities of independent, combined, and conditional events B calculating and interpreting expected values and standard deviations of discrete random variables C applying distributions such as the Poisson, binomial, and normal

91585

Apply probability concepts in solving problems

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