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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:
[tex]P(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}[/tex]
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:
[tex]P(1)=\frac{4!}{1!(4-1)!}*0.4^{1}*(1-0.4)^{4-1}=0.3456[/tex]
[tex]P(2)=\frac{4!}{2!(4-2)!}*0.4^{2}*(1-0.4)^{4-2}=0.3456[/tex]
[tex]P(3)=\frac{4!}{3!(4-3)!}*0.4^{3}*(1-0.4)^{4-3}=0.1536[/tex]
[tex]P(4)=\frac{4!}{4!(4-4)!}*0.4^{4}*(1-0.4)^{4-4}=0.0256[/tex]
Finally, P is equal to:
P = 0.3456 + 0.3456 + 0.1536 + 0.0256
P = 0.8704
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 is 0.8704.
Given :
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.
According to the Binomial distribution the probability is given by :
[tex]\rm P(x) = \dfrac{n!}{x!(n-x)!}\times p^x \times (1-p)^{n-x}[/tex]
where n is the event and x is the success.
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 is:
[tex]\rm P = P(1) + P(2) + P(3) + P(4)[/tex]
Here, n = 4 and P = 0.4.
[tex]\rm P(1)= \dfrac{4!}{1!(4-1)!}\times 0.4^1 \times (1-0.4)^{4-1} = 0.3456[/tex]
[tex]\rm P(2)= \dfrac{4!}{2!(4-2)!}\times 0.4^2 \times (1-0.4)^{4-2} = 0.3456[/tex]
[tex]\rm P(3)= \dfrac{4!}{3!(4-3)!}\times 0.4^3 \times (1-0.4)^{4-3} = 0.1536[/tex]
[tex]\rm P(4)= \dfrac{4!}{1!(4-4)!}\times 0.4^4 \times (1-0.4)^{4-4} = 0.0256[/tex]
Now, the value of P is given by:
[tex]\rm P=P(1)+P(2)+P(3)+P(4)[/tex]
P = 0.3456 + 0.3456 + 0.1536 + 0.0256
P = 0.8704
For more information, refer to the link given below:
https://brainly.com/question/23017717
