- #1
kliker
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Homework Statement
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?
Homework Equations
BAYES
P(A/B) = P(A AND B)/P(B) = P(A)*P(B/A)/P(B)
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 I am doing wrong?
thanks in advance
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