Need help in part b and c please urgently

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There are 4 possible outcomes where H=hard, E=easy: HH, HE, EH, EE.

The first letter is the type of exam, the second letter is the type of last question.

The exam probabilities are H=0.80, E=1-0.80=0.20.

When exam=H, the question probabilities are H=0.90, E=0.10;

When exam=E, the question probabilities are H=0.15, E=0.85.

So the combined probabilities for the 4 outcomes are the products of the probabilities:

HH=0.8×0.9=0.72

HE=0.8×0.1=0.08

EH=0.2×0.15=0.03

EE=0.2×0.85=0.17

These combined probabilities sum to 1, which means we've covered all outcomes.

a) 0.20 (as you probably knew)

b) HE+EE=0.08+0.17=0.25

c) HH+EH=0.72+0.03=0.75 tells us the probability of the last question being hard, but BEFORE taking the exam. But this is a given (AFTER the exam) so we need to know the relative probability, GIVEN the last question was hard, of EH/(HH+EH)=0.03/0.75=0.04.
by Top Rated User (982k points)

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