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A professor gives a true-false exam consisting of thirty T-F questions. The questions whose answers are “true” are randomly distributed among the thirty questions. The professor thinks that 3/4 of the class are serious, and have correctly mastered the material, and that the probability of a correct answer on any question from such students is 75%. The remaining students will answer at random. She glances at two questions from a test picked haphazardly. Both questions are answered correctly. What is the probability that this is the test of a serious student?

Using the Bayes theorem

$$\Pr(A|B) = \frac{\Pr(A)\Pr(B|A)}{\Pr(A)\Pr(B|A)+\Pr(A^c)\Pr(B|A^c)} = \frac{.75*.75}{(.75*.75)+(.25*.5)} = .818$$

I think I am on the right track. However, I am unsure if I need to do something to account for the fact that 2 questions were answered correctly? Any thoughts?

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  • $\begingroup$ What are the events $A,B$ here? $\endgroup$ – JPi Apr 26 '14 at 1:54
  • $\begingroup$ event A is the test belongs to a serious student and event B is getting an answer correct on the exam $\endgroup$ – Brian Apr 26 '14 at 1:57
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$B$ should be getting two answers correctly. So

$$ \Pr(A|B) = \frac{.75 \times .75^2}{.75 \times .75^2 + .25 \times .5^2}.$$

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    $\begingroup$ Thank you I greatly appreciate the help $\endgroup$ – Brian Apr 26 '14 at 2:20
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Let $A$ be the event the student is "serious" and $B$ the event she answers $2$ randomly chosen questions correctly. We want $\Pr(A|B)$, which is $\frac{\Pr(A\cap B)}{\Pr(B)}$.

We first calculate $\Pr(B)$. The event $B$ can happen in two ways: (i) the student is serious, and answers the two questions correctly and (ii) the student is not serious, but answers correctly.

The probability of (i) is $(3/4)(0.75)^2$. The probability of (ii) is $(1/4)(0.5)^2$.

Add, and note that $\Pr(A\cap B)$ is just the probability of (i), which we already have computed.

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