###### Calculate probabilities using combinat ...

The Multiplication Principle of Counting Assume that you are conducting an experiment where... **Read More**

\(A\) 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.

If \(A\) 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:

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

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 \(A\cup B=S\).

Consider 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,

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

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 3^{rd} roll,” then we can define event A’ as “rolling a 1 before the 3^{rd} roll” which is the same as rolling a 1 on either the 1^{st} or 2^{nd} roll.

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

The probability of rolling a 1 on the 2^{nd} roll is \({5}/{6}\) (because if we rolled a 1 on the 1^{st} 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. **