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A student answers a multiple choice examination with questions that have four possible answers each. Suppose that the probability that the student knows the answer to a question is 0.80 and the probability that the student guesses is 0.20. If the student guesses, the probability of guessing the correct answer is 0.25. The questions are independent, that is, knowing the answer on one question is not influenced by the other question.

(a) If there is one question on the exam and he answered the question correctly, what is the probability he knew the answer?

(b) If there are two questions on the exam and he answered both questions correctly, what is the probability he knew both answers?

(c) How would you generalize the above to n questions, that is, if the student answered an infinite number of questions correctly, what is the probability he knew the answers?

I know the answer to A using Bayes Theorem is

$\ P(A∣C)=\frac{P(C∣A)P(A)}{P(C∣A)P(A)+P(C∣Ac)P(Ac)}$

$\ \frac{(.8)(1)}{(.8)(1)+(.25)(.20)}$

But I'm completely stuck on B and C.

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  • $\begingroup$ I am not sure I would agree with "the probability of guessing correctly goes down as more and more questions are answered correctly" in your title $\endgroup$ – Henry Jan 26 at 17:59
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $\checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?. $\endgroup$ – NCh Feb 15 at 15:42
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HINTS

For part c), the hardest part is computing the probability that he answers all $n$ questions correctly. Suppose he knows the answers to $k$ questions. There are $\binom{n}{k}$ ways to pick the questions, and the probability that he knows the answer to exactly those questions is $.8^n\cdot.2^{n-k}$. The probability that hr guesses correctly on all the questions he doesn't know is $.25^{n-k}$. All together, the probability that he answers all questions correctly is $$\sum_{k=0}^n\binom{n}{k}.8^k(.2\cdot.25)^{n-k}=\sum_{k=0}^n\binom{n}{k}.8^k\cdot.05^{n-k}$$

Do you see how to simplify this? Can you answer part c) now?

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  • $\begingroup$ Alternatively, the probability he answers a particular question correctly is $0.85$ and, since each question's response is independent, the probability he answers them all correctly is $0.85^n$. Similarly, given he answered a particular question correctly, the probability he knew the answer was $\frac{0.8}{0.85}$, and so if he answered all $n$ questions correctly the probability he know them all is $\left(\frac{0.8}{0.85}\right)^n$ $\endgroup$ – Henry Jan 26 at 17:56
  • $\begingroup$ @Henry I thought of the argument in the second sentence of your comment, but I wasn't sure it was rigorous. Can we always multiply contingent probabilities like this? (I should have thought of the argument in the first sentence, which is clearly okay.) $\endgroup$ – saulspatz Jan 26 at 18:05
  • $\begingroup$ saulspatz: I would have thought you can multiply if they are independent. In one sense that is the definition of independent so is tautological, but in another sense information about whether one response was knowledge or guess here would not provide information about whether another response was knowledge or guess $\endgroup$ – Henry Jan 26 at 18:20
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Let $A_1$, $A_2$ be the events that student knows 1st and 2nd questions correspondingly. And let $C_1$, $C_2$ be the events that he gave right answers to these questions.

We need to find $$ \mathbb P(A_1\cap A_2\mid C_1 \cap C_2) = \frac{\mathbb P(A_1\cap A_2 \cap C_1\cap C_2)}{\mathbb P(C_1\cap C_2)}=\frac{\mathbb P((A_1\cap C_1) \cap (A_2\cap C_2))}{\mathbb P(C_1)\mathbb P(C_2)} $$ (since $A_1\cap C_1$ and $A_2\cap C_2$ are independent) $$=\frac{\mathbb P(A_1\cap C_1)\cdot \mathbb P(A_2\cap C_2)}{\mathbb P(C_1)\mathbb P(C_2)} = \mathbb P(A_1\mid C_1) \times \mathbb P(A_2\mid C_2) $$ Both this probabilities are from (a). So the probability to (b) is squared probability from (a), and for (c) it is $n$-th power.

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