*17. Given PO)-0.339, p(Z)=0315, and that events Yanaz are disjoint, find P(YorZ

Question

*17. Given PO)-0.339, p(Z)=0315, and that events Yanaz are disjoint, find P(YorZ) 0 ( For questions "18 through "22, round your answers to three decimal places (3 points each) Show work 18. A charity fundraises by calling a list of prospective donors to ask for pledges. It is able to reach 42.7% of the people on its list. Of those reached 32.3% make a pledge, what is the probability that a randomly chosen prospective donor is reached and the donor makes a pledge? and f . : 4477 .323 : . I3742 1 2/. 138 19 Of new motor vehicles sold to individuals in a recent year, 86.9% were domestic ma light trucks (where the term "light trucks" refers to SUVs and minivans), 788% domestic makes, and 53.5% were light trucks. What is the probability that a randomly selected new motor vehicle sold to an individual is a domestic light truck? kes or 20. The probability that steroid test A will detect the presence of steroids in a steroid user is 0.945. The probability that steroid test B will detect the presence of steroids in a steroid user is 0.910. Given that the two tests operate independently of each other, what is the probability that both of them will detect the presence of steroids in a steroid user?

Explanation / Answer

Question 17:

Here we are given that P(Y) = 0.339, P(Z) = 0.315

As Y and Z are disjoint, therefore P(Y and Z ) = 0

Using law of addition of probability, we get:

P(Y or Z ) = P(Y) + P(Z) - P(Y and Z) = 0.339 + 0.315 - 0 = 0.654

Therefore 0.654 is the required probability here.

Question 18:

Here we are given that:

P( reached ) = 0.427, P( pledge | reached ) = 0.323

Now the required probability is computed using Bayes theorem as:

P( reached and pledged ) = P( pledge | reached )P( reached ) = 0.323*0.427 = 0.1379

Therefore 0.1379 is the required probability here.

Question 19:

Here we are given that:

P( domestic or light trucks ) = 0.869,
P( domestic ) = 0.788 and P( light trucks ) = 0.535

Therefore, using law of addition of probability, we get:

P( domestic and light trucks ) = P( domestic ) + P( light trucks ) - P( domestic or light trucks )

P( domestic and light trucks ) = 0.788 + 0.535 - 0.869 = 0.454

therefore 0.454 is the required probability here.

Question 20:

Here, we are given that

P( A positive | steroid ) = 0.945,
P( B positive | steroid ) = 0.910

Now as the events are independent, therefore:

P( both positive | steroid ) = P( A positive | steroid )P( B positive | steroid ) = 0.945*0.91 = 0.85995

Therefore 0.85995 is the required probability here.

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