A multiple choice test contains $10$ questions, each with $3$ possible answers (of which only one is correct). If a student answers each question by rolling a die and choosing the first answer if the die shows $1$ or $2$, the second answer if the die shows $3$ or $4$, or the third answer if the dies shows $5$ or $6$, what is the probability that the student will get exactly $6$ correct answers?
I know that I must use a binomial distribution. I know that $n = 10$ trials, and I define $X$ to be the number of questions answered correctly in $10$ trials. Because one in three answers is correct, the probability of answering correctly would be $1/3$. But then the die rolling affects this probability, which is what I'm not sure how to figure. For example, the probability of choosing the first answer is $1/3$, and the probability of the first answer being the correct answer is $1/3$, but how do I combine these probabilities?
Do not worry, it turns out that it is just $1/3$.
We proceed mechanically. Consider a particular question. Let the event $C_i$ be the event that option $i$ is correct, let $R_i$ be the event that you roll $i$. Then \begin{align*} P(\text{Correct}) &= P(C_1R_1)+P(C_2R_2)+P(C_3R_3)\\ &= P(C_1)P(R_1)+P(C_2)P(R_2)+P(C_3)P(R_3)\\ &= \frac{1}{3}\frac 13+\frac13\frac13+\frac13\frac13\\ &=3\cdot\frac19\\ &=\frac{1}{3} \end{align*} where the first equality is true by the law of total probability and the second equality is true by assuming independence between your guess and the answer. Hence, the chance to be correct is $1/3$.
There other ways to arrive at this conclusion.