Question

A police officer is using a radar device to check motorists’ speeds. Prior to beginning the speed check, the officer estimates that 40 percent of motorists will be driving more than 5 miles per hour over the speed limit. Assuming that the police officer’s estimate is correct, what is the probability that among 4 randomly selected motorists, the officer will find at least 1 motorist driving more than 5 miles per hour over the speed limit?

Answer 1

Answer:

The probability is 0.8704

Step-by-step explanation:

A Binomial distribution apply when we have n identical and independent events with two possible results, success or fail, and a probability p of success and 1-p of fail. Then, the probability that from the n events, x are success is:

$P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}$

Then, the number of motorist driving more than 5 miles per hour over the speed limit follow a Binomial Distributions,  because we have 4 randomly selected motorists with two possible options: they are driving more than 5 miles per hour over the speed limit or they aren't, and the probability that they are driving more that 5 miles per hour over the speed is 0.4.

Finally, the probability P that among 4 randomly selected motorists, the officer will find at least 1 motorist driving more than 5 miles per hour over the speed limit is:

P = P(1) + P(2) + P(3) + P(4)

Where n is equal to 4 and p is equal to 0.4. Replacing the values for every x, we get:

$P(1)=\frac{4!}{1!(4-1)!}*0.4^{1}*(1-0.4)^{4-1}=0.3456$

$P(2)=\frac{4!}{2!(4-2)!}*0.4^{2}*(1-0.4)^{4-2}=0.3456$

$P(3)=\frac{4!}{3!(4-3)!}*0.4^{3}*(1-0.4)^{4-3}=0.1536$

$P(4)=\frac{4!}{4!(4-4)!}*0.4^{4}*(1-0.4)^{4-4}=0.0256$

Finally, P is equal to:

P = 0.3456 + 0.3456 + 0.1536 + 0.0256

P = 0.8704