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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.

1 Answers 1

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Hints for both as below. You should be able to work it out.

$b)\mathop{\mathbb{E}}[X] = \mathop{\mathbb{E}}[A] - \dfrac{\mathop{\mathbb{E}}[F]}{3}$

$Var[X] = Var[A] + \dfrac{Var[F]}{9} - \dfrac{2Cov(A,F)}{3} $

$c) P(X\geq5) = 1 - P(X<5)$

  • 0
    Could you explain why they are independent $A$ and $F$? Thank you very much. – 2017-02-14
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    Of course I understand now. Another question: Why is $P(X \geq 5) = 1 - P(X \leq 5)$? I thought it was $P(X \geq 5) = 1 - P(X < 4)$ – 2017-02-14
  • 0
    It's not. I have just edited. – 2017-02-14
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    And $1 - P(X < 5) = 1 - P(X ≤ 4)$? Thanks. – 2017-02-14
  • 1
    Correct, this is an alternative way of writing it. – 2017-02-14