Homework Help: What am I doing wrong? (Probability)

1. May 17, 2010

kliker

1. The problem statement, all variables and given/known data

the probability for a student to answer correctly in one question when he has studied is 0.8 and when he hasnt studied is 0.3. In order to study he throws a dice, if the result is 1 he won't study. At the exam, 2 out of 3 questions were correct. What's the probability that the student has studied?

2. Relevant equations

BAYES

P(A/B) = P(A AND B)/P(B) = P(A)*P(B/A)/P(B)

3. The attempt at a solution

i find the probability for him to answer 1 question correctly

C = {He answers 1 question correctly}
S = {The student has studied}
S' = {The student hasnt studied}

P(C) = P(C/S)*P(S) + P(C/S')*P(S') = 0.8*5/6+0.3*1/6=0.716

now i find the probability that he answers 2 questions correctly

2C = {He answers 2 questions correctly}

P(2C) = C(3,2)*(0.716)^2*(1-0.716) = 0.436788

hence

P(S/2C) = P(S AND 2C)/P(2C) (2)

we know P(2C) hence we need to find P(S AND 2C)

we use bayes

P(S AND 2C) = P(S)*P(2C/S) = 5/6*P(2C/S) (1)

P(2C/S) is the probability that he answers 2 questions correctly if we know that he has studied

hence P(2C/S) = C(3,2)*(0.8)^2*0.2 = 0.384

So (1) = 0.32

hence the probability that he has studied if we know that he answered 2 questions correctly is from (2) => 0.32/0.436788 = 0.732621

the correct answer is 0.9104 though

can someone explain me what im doing wrong?