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Suppose there are two multiple choice questions with 4 choices each. Assume you answer the first question by choosing one of the four answer uniformly at random. You answer the second questions by choosing, again uniformly at random, one of the three answers you did not choose in the first question. What is the probability that you get the second question correctly?

The probability of getting question 2 correct seems dependent of what answer we choose for question 1.

Question 1: $\frac{1}{4}$

Question 2: $\frac{1}{3}$

Would the the probability of getting question 2 correct be $\frac{1}{4} \cdot \frac{1}{3} = \frac{1}{12}$?

How can we consider the possibility of the MC answer pick from the first question was the correct choice for the second question? Therefore making the probability of getting the second question zero.

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  • $\begingroup$ Do you have four answers to share among the two questions, or four answers for each question? Your first sentence seems to indicate four for each question. If so, do you have reason to believe they do not have the same answer? $\endgroup$
    – Ross Millikan
    Apr 17, 2014 at 3:53

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You can also use casework to verify:

Case 1: The answer to Question 1 is correct for Question 2

This happens with probability $1/4$. There is no way we can get Question 2 correct, so the probability in this case is $1/4$*$0$=$0$.

Case 2: The answer to Question 1 is incorrect for Question 2

This happens with probability $3/4$. Thus the correct answer for Question 2 must be in the remaining 3, which we will get right on a $1/3$ chance: $3/4$*$1/3$=$1/4$.

Thus the probability we get Question 2 right is $0$+$1/4$=$1/4$.

It should make sense that they are independent, because every single answer for Question 2 has an equal chance of being picked solely due to the fact that every single answer for Question 1 has an equal chance of being picked.

I believe that is what you're asking; correct me if I misinterpreted the question.

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There are $12$ ordered pairs $(a_1,a_2)$ consisting of answers selected for the first and second questions respectively (since $a_1\neq a_2$). Each of these will be selected with probability $\frac{1}{12}$. The correct answer for question two will be the second element of exactly $3$ of these ordered pairs. Hence the probability of getting the second question correct is $\frac{3}{12} = \frac{1}{4}$. Note that this is the same as if we had just chosen our second answer independent of the first.

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