In Exercises 9–16, use the Poisson distribution to find the indicated probabilities.

Car Fatalities The recent rate of car fatalities was 33,561 fatalities for 2969 billion miles traveled (based on data from the National Highway Traffic Safety Administration). Find the probability that for the next billion miles traveled, there will be at least one fatality. What does the result indicate about the likelihood of at least one fatality?

It is given that out of 2969 billion miles traveled, a total of 33561 fatalities have occurred.

The number of fatalities is equal to 33561.

The number of billion miles traveledis equal to 2969.

The mean number of fatalities per billion miles traveledis computed below:

\(\begin{aligned}{c}\mu = \frac{{{\rm{Number}}\;{\rm{of}}\,{\rm{fatalities}}}}{{{\rm{Number}}\;{\rm{of}}\;{\rm{billion}}\;{\rm{miles}}\;{\rm{travelled}}}}\\ = \frac{{33561}}{{2969}}\\ = 11.30\end{aligned}\)

Thus, the mean number of fatalities per billion miles traveledis equal to 11.30.

Let X be the number of fatalities per billion miles traveled.

Here, X follows a Poisson distribution with mean equal to\({\kern 1pt} \mu = 11.30\).

The formula of Poisson distribution is:

\[P\left( {X = x} \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\]

The probability that the number of fatalities per billion miles traveledwill be at least equal to 1 is computed below;

\[\begin{aligned}{c}P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\ = 1 - P\left( 0 \right)\\ = 1 - \frac{{{{\left( {11.30} \right)}^0}{{\left( {2.71828} \right)}^{ - 11.30}}}}{{0!}}\\ = 1 - 0.0000124\end{aligned}\]

\[ = 0.9999876\]

Therefore, theprobability that the number of fatalities per billion miles traveledwill be at least equal to 1 is equal to 0.9999876.

Since the probability is approximately equal to 1, it is highly likely that at least 1 fatality will occur in the next billion miles traveled.