If $X=A-F/3$, how to calculate $E(X)$, $Var(X)$ and $P(X≥5)$?

Question

The exercise

An examination of questions with multiple answers, has 20 questions, and each question consists of 4 alternatives, one of which is correct.

The student's score is a random variable $ X $ given by $ X = A-\dfrac{F}{3} $,

where $A$ is the random variable "number of hits" and $F$ is the random variable "number of failures".

If a student answers at random all the questions:

• a) What is the distribution of variable $A$?
• b) What is the expectation and variance of $X$?
• c) What is the probability that the student will get at least $5$ points in the exam?

What I did

a) $A$~$B(20,1/4)$

I assumed that $F$~$B(20,3/4)$, so that:

• $E(A)=(20)*(1/4)=5$
• $E(F)=(20)*(3/4)=15$
• $Var(A)=(20)*(1/4)*(1-(1/4))=15/4$
• $Var(F)=(20)*(3/4)*(1-(3/4))=15/4$

If part (a) is ok (maybe not). How can I resolve parts (b) and (c)? Thank you very much.

Answer 1

There's less here than meets the eye; most of the question is based on the sleight of hand introduction of the variable $F$, which is just $A-20$ (These are NOT independent variables, just a trap for the unwary.) Plug that into the definition of $X$, and calculate away! Hint: $\Pr(X \geq 5) = \Pr(A > 8)$.