Need to find help in solving probability question

Question

. A professor gives only two types of exams and designs only two types of questions, “easy” and “hard”.

You will get a hard exam with probability 0.80. The probability that the last question on the exam is hard

is 0.90 if the exam is hard; the probability that the last question on the exam is hard is 0.15 if the exam is

easy.

(a)Before you attend an exam, estimate what is the probability that the last question on your exam will

be easy.

(b) Now you have attended the exam. The last question on the exam turned out to be hard. What is the

probability that you have attended an easy exam?

Answer 1

H=hard exam; E=easy exam

L=last question hard; ~L=last question easy

P(H)=.8 and P(E)=.2

P(L|H)=.9 and P(~L|H)=.1

P(L|E)=.15 and P(~L|E)=.85

P(~L and H)=P(H)*P(~L|H)=.8*.1=.08

P(~L and E)=P(E)*P(~L|E)=.2*.85=.17

Therefore P(~L)=.25 and P(L)=.75

P(L and E)=P(E)*P(L|E)=.2*.15=.03

P(E|L)=P(L and E)/P(L)=.03/.75=.04