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.

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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.

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

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Independent Events

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

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

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

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

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@mmerevise

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 $\color{red}A$ 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}

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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 LevelAQAEdexcelOCR

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

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.

Gold Standard Education

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

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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.

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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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