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The question is the following: Professor May B. Right often has her facts wrong, and answers each of her students' questions incorrectly with probability $1/4$, independent of other questions. In each lecture, May is asked $0$, $1$, or $2$ questions with equal probability $1/3$. Let $X$ and $Y$ be the number of questions May is aked and the number of questions she answers wrong in a given lecture, respectively. Find $P_{X,Y}{(2,1)}$.

So, I understand that we need to find $\text{P}((\text{She gets asked 2 questions})\cap(\text{She answers exactly one of them incorrectly}))$ $= 1/3*(1/4*3/4) = 3/48$, right? Since the event that she answers exactly one of the two questions incorrectly corresponds to the intersection of the events that she answers one question wrong and the other one right. But the book says that the answer should be $6/48$. Can someone please explain to me what I am getting wrong here? Thanks

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    I have modified the notation slightly to be more in line with the way a probability mass function is usually written. – 2017-02-18

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HINT: in the case where she is asked two questions, we have three possibilities for her answers: both correct with probability $(\frac {3}{4})^2=\frac {9}{16} \,\,\,$, one correct and one wrong with probability $2 \cdot \frac { 3}{4} \cdot \frac {1}{4}=\frac {6}{16} \,\,\,$, and both wrong with probability $(\frac {1}{4})^2=\frac {1}{16} \,\,\,$.

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    I understand the cases for both correct and both wrong, but why do we have to multipliply by 2 in the case for one correct and one wrong? Where does this factor come from? Thanks – 2017-02-18
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    Because you have two cases: the first is correct and the second is wrong, and vice versa. – 2017-02-18