Question

The National Safety Council publishes information about automobile accidents in Accident Facts. According to that document, the probability is .40 that a traffic fatality involves an intoxication or alcohol-impaired driver or nonoccupant. In eight traffic fatalities, find the probability that the number which involve an intoxicated or alcohol-impaired driver or nonoccupant is

Answer 1

Answer:

$P(X = x) = C_{8,x}.(0.4)^{x}.(0.6)^{8-x}$

In which x is the number of which we want to find the probability.

Step-by-step explanation:

For each traffic fatality, there are only two possible outcomes. EIther it involved an intoxicated or alcohol-impaired driver or nonoccupant, or it didn't. Traffic fatalities are independent of other traffic fatalities, which means that the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The probability is .40 that a traffic fatality involves an intoxication or alcohol-impaired driver or nonoccupant.

This means that $p = 0.4$

Eight traffic fatalities

This means that $n = 8$

Find the probability that the number which involve an intoxicated or alcohol-impaired driver or nonoccupant is

This is P(X = x), in which x is the number of which we want to find the probability. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = x) = C_{8,x}.(0.4)^{x}.(0.6)^{8-x}$