A management student has one more class to register for in t

A management student has one more class to register for in the upcoming semester. She is considering taking a class in Finance, Psychology, or Statistics. The probability that she will take a Finance class is 0.5, probability for Psychology is 0.1. The student’s advisor then gives the student estimated probabilities of passing in each class. If the student takes Finance, she will pass with probability 0.68. Her passing probability for Psychology is 0.91 and is 0.83 for Statistics.

a) Draw a well-labeled tree diagram to represent this problem.

b) What is the probability that the student failed and did not take Psychology?

c) What is the probability that the student passes this final class for the semester?

d) Suppose it is known that the student has failed the class; what is the probability that she took Statistics?

e) At the end of the semester it is known that the student passed the class. What is the probability that the class taken was Finance?

Solution

probability of considering class of Finance p(F)=0.5 ,

probability of considering class of Psychology p(P)=0.1,

probability of considering class of Statistics p(S)=1-0.5-0.1=0.4,

probability of passing and fail in Finance, Psychology and Statistics is given by respectively

p(pass/F)=0.68, p(fail/F)=0.0.32

p(pass/P)=0.91,p(fail/P)=0.09

p(pass/S)=0.83,p(fail/S)=0.17

probability of passsing=p(pass)=0.5*0.68+0.1*0.91+0.4*0.83=0.763

probability of failing =p(fail)=0.5*0.32+0.1*0.09+0.4*0.17=0.237

answer of part(a)

difficult to draw tree diagram on this platform.

answer of part(b)

p(S)*p(fail/S)+p(F)*p(fail/F)=0.4*0.17+0.5*0.32=0.068+0.160=0.228

answer of part(c)

probability of passsing=p(pass)=0.5*0.68+0.1*0.91+0.4*0.83=0.763

answer of part(d)

p(fail/S)=(0.4*0.17)/(0.5*0.32+0.1*0.09+0.4*0.17)=0.068/0.237=0.2869

answer of part (e)

p(pass/F)=0.1*0.91/(0.5*0.68+0.1*0.91+0.4*0.83)=0.091/0.763=0.1193