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?
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 $b=-1.5$.