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We consider a multiple-choice test with exactly three yes/no-questions. The following rating scheme is given: correct answer +1 point, wrong/missing answer -0.5 points - a negative result is NOT rounded up to 0. Now assume a student who is trying to guess all answers and who is ticking at most one possible answer. The probability that the student leaves one question is unchecked is 1/5 while the probability of ticking one of the two other answers is 2/5. Let A denote a random variable representing the points one student has collected. [...]

Determine parameters $a,b\in\mathbb{R}$ for $A=aX+b$ with $X\sim \text{Bin}(3, 2/5)$ and determine $\mathbb{E}[A]$.

(horribly translated assignment)

The only things I know right now, is that with $$\mathbb{E}[A]=\mathbb{E}[aX+b]=a\cdot\mathbb{E}[X]+b=a\cdot\left(3\cdot\frac{2}{5}\right)+b$$ I could do some magic, but I don't exactly know how to include the possible results (-1.5, 0, 1.5, 3) into the parameters $a$ and $b$.

Could someone give me some hints how to solve this?

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Hint: The student receives one point for a correct answer and $-1/2$ for an incorrect or blank answer and he gets $X$ right (getting $+1$ for each), while $3-X$ are marked incorrect or left blank (getting $-1/2$ for each). Hence, $A$, the student's score on the exam is given by .... – Dilip Sarwate May 19 '12 at 21:42
@DilipSarwate: Thats a really good hint (almost a full solution!). Shame on me why I haven't understood it before. This should result in $\mathbb{E}[A]=0.3$, shouldn't it? – Christian Ivicevic May 19 '12 at 21:49
up vote 2 down vote accepted

You want a linear transformation to change $(0,1,2,3)$ into $(-1.5,0,1.5,3)$.

The gaps are $1$ in the first case and $1.5$ in the second. So $a=\frac{1.5}{1} =1.5$.

Multiplying by $1.5$ changes $(0,1,2,3)$ into $(0,1.5,3,4.5)$. To get from $(0,1.5,3,4.5)$ to $(-1.5,0,1.5,3)$ you have to subtract $1.5$ from each term. So $b=-1.5$.

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Hint: Let $X_i=1$ if Question $i$ is answered correctly, and $X_i=0$ otherwise. Then the number of marks $M_i$ earned on Question $i$ is $1.5X_i -0.5$. (Check that this is correct in both the case $X_i=1$ and $X_i=0$.)

The total mark is $M=M_1+M_2+M_3$.

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