Mutually Exclusive Events and Independence

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Mutually Exclusive Events, Independence and Modelling

Mutually exclusive events are events that cannot both happen. Independent events are events that do not affect each other.

Also on this page we will discuss the assumptions made when making a probability model.

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Mutually Exclusive Events

Two events Acolor{red}A and Bcolor{blue}B are mutually exclusive if they can’t both happen – i.e. P(A∩B)=0mathbb{P}(color{red}Acolor{grey}capcolor{blue}Bcolor{grey})=0.

If Acolor{red}A and Bcolor{blue}B are mutually exclusive, then the probability of Acolor{red}A or Bcolor{blue}B is the sum of the probabilities of the individual events.

P(A∪B)=P(A)+P(B)mathbb{P}(color{red}Acolor{grey}cupcolor{blue}Bcolor{grey})=mathbb{P}(color{red}Acolor{grey})+mathbb{P}(color{blue}Bcolor{grey})

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Independent Events

If two events Acolor{red}A and Bcolor{blue}B are independent, then they do not affect each other. This means:

P(A∣B)=P(A)mathbb{P}(color{red}Acolor{grey}|color{blue}Bcolor{grey})=mathbb{P}(color{red}Acolor{grey})

P(B∣A)=P(B)mathbb{P}(color{blue}Bcolor{grey}|color{red}Acolor{grey})=mathbb{P}(color{blue}Bcolor{grey})

P(A∩B)=P(A)P(B)mathbb{P}(color{red}Acolor{grey}capcolor{blue}Bcolor{grey})=mathbb{P}(color{red}Acolor{grey})mathbb{P}(color{blue}Bcolor{grey})

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Modelling Assumptions

Most probability models have made several assumptions in order to be practical to use.

For example, when you roll a die, you assume that each number is equally likely. But in reality, the die could be biased, or could have a small probability of landing on its edge.

Common assumptions include:

  • Equally likely events: Are these events actually equally likely? Could there be other low probability outcomes? Are they significant enough to render the model redundant?
  • Modelling based on past data: Can we trust the data? How was it sampled? Is it relevant to what we are doing?
  • Randomness: Is the outcome genuinely random? Are we doing something that could affect the outcome without us knowing?
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Example 1: Mutually Exclusive Events

Events Acolor{red}A and Bcolor{blue}B are mutually exclusive. If P(A∪B)=0.9mathbb{P}(color{red}Acolor{grey}cupcolor{blue}Bcolor{grey})=0.9 and P(A)=0.3mathbb{P}(color{red}Acolor{grey})=0.3, what is the probability of Bcolor{blue}B?

[2 marks]

P(A∪B)=P(A)+P(B)mathbb{P}(color{red}Acolor{grey}cupcolor{blue}Bcolor{grey})=mathbb{P}(color{red}Acolor{grey})+mathbb{P}(color{blue}Bcolor{grey})

0.9=0.3+P(B)0.9=0.3+mathbb{P}(color{blue}Bcolor{grey})

P(B)=0.9−0.3=0.6begin{aligned}mathbb{P}(color{blue}Bcolor{grey})&=0.9-0.3[1.2em]&=0.6end{aligned}

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Example 2: Independent Events

Suppose P(A)=0.9mathbb{P}(color{red}Acolor{grey})=0.9, P(B)=0.4mathbb{P}(color{blue}Bcolor{grey})=0.4 and P(A∩B)=0.3mathbb{P}(color{red}Acolor{grey}capcolor{blue}Bcolor{grey})=0.3. Are Acolor{red}A and Bcolor{blue}B independent?

[1 mark]

For independent events, P(A)P(B)=P(A∩B)mathbb{P}(color{red}Acolor{grey})mathbb{P}(color{blue}Bcolor{grey})=mathbb{P}(color{red}Acolor{grey}capcolor{blue}Bcolor{grey})

Here, P(A)P(B)=0.9×0.4=0.36mathbb{P}(color{red}Acolor{grey})mathbb{P}(color{blue}Bcolor{grey})=0.9times 0.4=0.36

And P(A∩B)=0.3mathbb{P}(color{red}Acolor{grey}capcolor{blue}Bcolor{grey})=0.3

So they are not equal and the events are not independent.

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Example 3: Assumptions in Modelling

Define a probability model for flipping a coin twice and state three assumptions you have made.

[5 marks]

We can define the model with a tree diagram:

We have assumed that:

  • Heads and tails are the only possible outcomes.
  • Both outcomes are equally likely.
  • The coin tosses are independent.
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Mutually Exclusive Events and Independence Example Questions

Question 1: P(A)=0.35mathbb{P}(A)=0.35, P(B)=0.4mathbb{P}(B)=0.4 and P(C)=0.3mathbb{P}(C)=0.3

Is it possible for the events AA, BB and CC to be mutually exclusive?

[2 marks]

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If the events are mutually exclusive, then P(A∪B∪C)=P(A)+P(B)+P(C)=0.35+0.4+0.3=1.05mathbb{P}(Acup Bcup C)=mathbb{P}(A)+mathbb{P}(B)+mathbb{P}(C)=0.35+0.4+0.3=1.05 which is a probability greater than 11, which is not allowed. Hence, the events cannot be mutually exclusive.

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Question 2: If we assume that rolling a dice and tossing a coin are independent of each other, what is the probability of rolling a 66 then flipping a tails.

[2 marks]

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For independent events, to find out the probability of both happening, we just multiply the probabilities of each – remember P(A∩B)=P(A)P(B)mathbb{P}(Acap B)=mathbb{P}(A)mathbb{P}(B)

 

The probability of rolling a 66 is 16dfrac{1}{6} and the probability of flipping a tails is 12dfrac{1}{2}

 

Hence, the probability of both is 16×12=112dfrac{1}{6}timesdfrac{1}{2}=dfrac{1}{12}

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Question 3: Previous data indicates that the probability of a train stopping at Harrogate train station in a five minute window is 16frac{1}{6}, and that each five minute window is independent of all of the others.

 

a) What is the probability of two trains arriving in ten minutes?

 

b) State an assumption you made about the data.

[2 marks]

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a) P(two trains in ten minutes)=16×16=136mathbb{P}(text{two trains in ten minutes})=dfrac{1}{6}timesdfrac{1}{6}=dfrac{1}{36}

b) We assumed:

  • The data is reliable and sampled correctly.
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Question 4: For two events AA and BB, we have P(A)=0.5mathbb{P}(A)=0.5, P(A∣B)=0.7mathbb{P}(A|B)=0.7 and P(A∩B)=0.15mathbb{P}(Acap B)=0.15.

 

a) Find:

i) P(A′)mathbb{P}(A')

ii) P(A′∣B)mathbb{P}(A'|B)

iii) P(B)mathbb{P}(B)

iv) P(B′)mathbb{P}(B')

v) P(A′∩B)mathbb{P}(A'cap B)

vi) P(A∪B)mathbb{P}(Acup B)

 

b) Are AA and BB independent?

[8 marks]

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a) i)

P(A′)=1−P(A)=1−0.5=0.5begin{aligned}mathbb{P}(A')&=1-mathbb{P}(A)[1.2em]&=1-0.5[1.2em]&=0.5end{aligned}

 

ii)

P(A′∣B)=1−P(A∣B)=1−0.7=0.3begin{aligned}mathbb{P}(A'|B)&=1-mathbb{P}(A|B)[1.2em]&=1-0.7[1.2em]&=0.3end{aligned}

 

iii)

P(B)=P(A∩B)P(A∣B)=0.150.7=0.214begin{aligned}mathbb{P}(B)&=dfrac{mathbb{P}(Acap B)}{mathbb{P}(A|B)}[1.2em]&=dfrac{0.15}{0.7}[1.2em]&=0.214end{aligned}

 

iv)

P(B′)=1−P(B)=1−0.214=0.786begin{aligned}mathbb{P}(B')&=1-mathbb{P}(B)[1.2em]&=1-0.214[1.2em]&=0.786end{aligned}

 

v) P(B)=P(A∩B)+P(A′∩B)mathbb{P}(B)=mathbb{P}(Acap B)+mathbb{P}(A'cap B)

 

0.214=0.15+P(A′∩B)0.214=0.15+mathbb{P}(A'cap B)

 

P(A′∩B)=0.214−0.15=0.064begin{aligned}mathbb{P}(A'cap B)&=0.214-0.15[1.2em]&=0.064end{aligned}

 

vi)

P(A∪B)=P(A)+P(B)−P(A∩B)=0.5+0.214−0.15=0.564begin{aligned}mathbb{P}(Acup B)&=mathbb{P}(A)+mathbb{P}(B)-mathbb{P}(Acap B)[1.2em]&=0.5+0.214-0.15[1.2em]&=0.564end{aligned}

 

b) If AA and BB are independent, then P(A)P(B)=P(A∩B)mathbb{P}(A)mathbb{P}(B)=mathbb{P}(Acap B).

P(A)P(B)=0.5×0.214=0.107mathbb{P}(A)mathbb{P}(B)=0.5times 0.214=0.107

P(A∩B)=0.15mathbb{P}(Acap B)=0.15

So they are not equal, so the events are not independent.

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Additional Resources

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Specification Points Covered

M1 – Understand and use mutually exclusive and independent events when
calculating probabilities

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