# Mutually Exclusive Events and Independence

## Mutually Exclusive Events and Independence Revision

**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**.

**Mutually Exclusive Events**

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

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

\mathbb{P}(\color{red}A\color{grey}\cup\color{blue}B\color{grey})=\mathbb{P}(\color{red}A\color{grey})+\mathbb{P}(\color{blue}B\color{grey})

**Independent Events**

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

\mathbb{P}(\color{red}A\color{grey}|\color{blue}B\color{grey})=\mathbb{P}(\color{red}A\color{grey})

\mathbb{P}(\color{blue}B\color{grey}|\color{red}A\color{grey})=\mathbb{P}(\color{blue}B\color{grey})

\mathbb{P}(\color{red}A\color{grey}\cap\color{blue}B\color{grey})=\mathbb{P}(\color{red}A\color{grey})\mathbb{P}(\color{blue}B\color{grey})

**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?

**Example 1: Mutually Exclusive Events**

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

**[2 marks]**

\mathbb{P}(\color{red}A\color{grey}\cup\color{blue}B\color{grey})=\mathbb{P}(\color{red}A\color{grey})+\mathbb{P}(\color{blue}B\color{grey})

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

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

**Example 2: Independent Events**

Suppose \mathbb{P}(\color{red}A\color{grey})=0.9, \mathbb{P}(\color{blue}B\color{grey})=0.4 and \mathbb{P}(\color{red}A\color{grey}\cap\color{blue}B\color{grey})=0.3. Are \color{red}A and \color{blue}B** independent**?

**[1 mark]**

For independent events, \mathbb{P}(\color{red}A\color{grey})\mathbb{P}(\color{blue}B\color{grey})=\mathbb{P}(\color{red}A\color{grey}\cap\color{blue}B\color{grey})

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

And \mathbb{P}(\color{red}A\color{grey}\cap\color{blue}B\color{grey})=0.3

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

**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.

## Mutually Exclusive Events and Independence Example Questions

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

Is it possible for the events A, B and C to be mutually exclusive?

**[2 marks]**

If the events are mutually exclusive, then \mathbb{P}(A\cup B\cup C)=\mathbb{P}(A)+\mathbb{P}(B)+\mathbb{P}(C)=0.35+0.4+0.3=1.05 which is a probability greater than 1, which is not allowed. Hence, the events cannot be mutually exclusive.

**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 6 then flipping a tails.

**[2 marks]**

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

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

Hence, the probability of both is \dfrac{1}{6}\times\dfrac{1}{2}=\dfrac{1}{12}

**Question 3:** Previous data indicates that the probability of a train stopping at Harrogate train station in a five minute window is \frac{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]**

a) \mathbb{P}(\text{two trains in ten minutes})=\dfrac{1}{6}\times\dfrac{1}{6}=\dfrac{1}{36}

b) We assumed:

- The data is reliable and sampled correctly.

**Question 4: **For two events A and B, we have \mathbb{P}(A)=0.5, \mathbb{P}(A|B)=0.7 and \mathbb{P}(A\cap B)=0.15.

a) Find:

i) \mathbb{P}(A')

ii) \mathbb{P}(A'|B)

iii) \mathbb{P}(B)

iv) \mathbb{P}(B')

v) \mathbb{P}(A'\cap B)

vi) \mathbb{P}(A\cup B)

b) Are A and B independent?

**[8 marks]**

a) i)

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

ii)

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

iii)

\begin{aligned}\mathbb{P}(B)&=\dfrac{\mathbb{P}(A\cap B)}{\mathbb{P}(A|B)}\\[1.2em]&=\dfrac{0.15}{0.7}\\[1.2em]&=0.214\end{aligned}

iv)

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

v) \mathbb{P}(B)=\mathbb{P}(A\cap B)+\mathbb{P}(A'\cap B)

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

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

vi)

\begin{aligned}\mathbb{P}(A\cup B)&=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\cap B)\\[1.2em]&=0.5+0.214-0.15\\[1.2em]&=0.564\end{aligned}

b) If A and B are independent, then \mathbb{P}(A)\mathbb{P}(B)=\mathbb{P}(A\cap B).

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

\mathbb{P}(A\cap B)=0.15

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

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