Showing the probability of guessing correctly goes down as more and more questions are answered correctly.

Question

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.

Answer 1

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?

Answer 2

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.