Abiometric security device using fingerprints erroneously refuse to admit 1 in 1,000 authorized person from a facility containing classified information. the device will erroneously admit 1 in 1,000,000 unauthorized persons. assume that 95 percent of those who seek access are authorized. if the alarm goes off and a person is refused admission, what is the probability that the person was really authorized?

0.01864

Step-by-step explanation:

Let A be the event that a person was authorized and NA be the event that the person was not authorized. Let R be the event that a person was refused admission and G be the event that a person was granted admission.

Then we have to find P(A|R) = [P(A)*P(R|A)]/P(R)

putting value we get

P(R) = P(A)P(R|A) + P(NA)P(R|NA) = 0.95*(1/1000) + 0.05*((1,000,000-1)/1,000,000) = 0.0.05094995

P(A|R) = [P(A)*P(R|A)]/P(R)

= [0.95*(1/1000)]/(0.05094995)

=0.01864

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation:

the answer is 9.8975

explanation:

when you multiply the "39.59" with the 25% it reduces it and shows its final discounted price. hope this !

step-by-step explanation: