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Calculate probabilities of mutually exclusive events

Calculate probabilities of mutually exclusive events

Mutually Exclusive Events

\(A\) and \(B\) are mutually exclusive events if \(A\) and \(B\) cannot both occur at the same time. For example, when a coin is flipped, it cannot land on both the head and tail simultaneously. Therefore, we consider the events “landing heads” and “landing tails” as mutually exclusive events.

If events are mutually exclusive, then their intersection \(\left( A\cap B \right) \) is equal to \(\emptyset\) (null set).

In the diagram below, \(A\) and \(B\) are mutually exclusive events.

Mutually Exclusive EventsIf \(A\) and \(B\) are mutually exclusive, then the probability of \(A\) or \(B\) occurring is the sum of the probability of event \(A\) occurring and the probability of event \(B\) occurring, as shown in the formula below:

$$ P(A \cup B) = P(A) + P(B) $$

Example: Mutually Exclusive Events

Consider drawing 2 cards from a standard deck. Let event \(A\) be drawing a Jack and event \(B\) be drawing a Queen. The probability of event \(A\) occurring is \({4}/{52}\) or \({1}/{13}\) and the probability of event \(B\) is also \({4}/{52}\) or \({1}/{13}\). Therefore, we can calculate:

$$ P(A\cup B)=P(A)+P(B)={ 1 }/{ 13 }+{ 1 }/{ 13 }=2/{ 13 } $$

Events \(A\) and \(B\) are exhaustive events if \(A\cup B\) is equal to the entire sample space, \(S\). It follows then that if \(A\) and \(B\) are mutually exclusive and exhaustive events then \(A\cap B=\emptyset \) and \(A\cup B=S\).

Exhaustive EventsConsider event \(A\) and its  complement  \(A^{c}\) (or \(A^{\prime}\)). These events are mutually exclusive and because they make up the entire sample space, \(S\), they are also exhaustive. Therefore,

$$ \begin{align*} & P\left( A \right) +P\left( { A }^{ ’ } \right) =1,\quad \text{and} \\ & P\left( A \right) =1-P\left( { A }^{ ’ } \right) \\ \end{align*} $$

Example: Mutually Exclusive Events

Consider the experiment rolling a 6-sided die until a 1 is rolled. If we define event \(A\) as “NOT rolling a 1 before the 3rd roll,” then we can define event A’ as “rolling a 1 before the 3rd roll” which is the same as rolling a 1 on either the 1st or 2nd roll.

The probability of rolling a 1 on the 1st roll is \({1}/{6}\).

The probability of rolling a 1 on the 2nd roll is \({5}/{6}\) (because if we rolled a 1 on the 1st roll we would have stopped) * \({1}/{6}\) = \({5}/{36}\)

$$ \begin{align*} & P\left( { A }^{ ’ } \right) ={ 1 }/{ 6 }+{ 5 }/{ 36 }={ 11 }/{ 36 } \\ & P\left( A \right) =1-P\left( { A }^{ ’ } \right) =1-{ 11 }/{ 36 }={ 25 }/{ 36 } \\ \end{align*} $$

Learning Outcome

1.d Topic: General Probability – Calculate probabilities of mutually exclusive events. 

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