Mutually Exclusive Example:

In the United States there are 59 different species of mammals that are endangered, 75 different species of birds that are endangered, and 68 species of fish that are endangered. If one of these animal is selected at random, find the probability that it is either a mammal or a fish.

Step 1:

Note these are mutually exclusive, that is no animal is both a mammal and a fish, so we can use the mutually exclusive addition rule:

\[P(\text{mammal or fish})=P(\text{mammal})+P(\text{fish})\]

Step 2:

Now we can calculate

\[P(\text{mammal})=\frac{\text{number of endangered mammals}}{\text{number of endangered animals}}=\frac{59}{202}\]

and

\[P(\text{fish})=\frac{\text{number of endangered fish}}{\text{number of endangered animals}}=\frac{68}{202}\]

so

\[P(\text{mammal or fish})=\frac{59}{202}+\frac{68}{22}=\frac{59+68}{202}\approx 0.63\]

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General Addition Rule Example:

In a hospital unit there are 8 nurses and 5 physicians; 7 nurses and 3 physicians are females. If a staff person is selected, find the probability that the subject is a nurse or a male.

Step 1:

We see in this example that a subject can indeed be a nurse and a male, so we will use the general addition formula:

\[P(\text{nurse or male})=P(\text{nurse})+P(\text{male})-P(\text{male nurse})\]

Step 2:

Now we calculate

\[P(\text{nurse})=\frac{\text{number of nurses}}{\text{total number of staff}}=\frac{8}{13}\]

and

\[P(\text{nurse})=\frac{\text{number of males}}{\text{total number of staff}}=\frac{3}{13}\]

and

\[P(\text{nurse})=\frac{\text{number of male nurses}}{\text{total number of staff}}=\frac{1}{13}\]

so

\[P(\text{nurse or male})=\frac{8}{13}+\frac{3}{13}-\frac{1}{13}=\frac{8+3-1}{13}=\frac{10}{13}\approx 0.77\]