# Mutually Exclusive Events and Independence

A LevelAQAEdexcelOCRAQA 2022

## 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.

A Level   ## 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.

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

A Level   ## Independent Events

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

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

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

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

A Level   ## 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?
A Level   ## 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}

A Level   ## 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.

A Level   ## 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.
A Level   ## Example Questions

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.

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 $\dfrac{1}{6}\times\dfrac{1}{2}=\dfrac{1}{12}$

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.

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.

A Level

A Level

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