6,974 views

An examination paper has $150$ multiple choice questions of one mark each, with each question having four choices. Each incorrect answer fetches $-0.25$ marks. Suppose $1000$ students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained by all these students is

1. $0$
2. $2550$
3. $7525$
4. $9375$

### 1 comment

Real-life application.

Answer is option D.

Probability of choosing the correct option = $\frac{1}{4}$
Probability of choosing a wrong option = $\frac{3}{4}$

So, expected mark for a question for a student = $\frac{1}{4} \times 1 + \frac{3}{4}\times (-0.25) = 0.0625$

Expected mark for a student for $150$ questions = $0.0625 \times 150 = 9.375$

So, sum total of the expected marks obtained by all $1000$ students = $9.375 \times 1000 = 9375$.

by

'Students choose all their answers randomly with uniform probability' Does this mean that it has unform distribution?
We want to calculate

$\bigcup_{i=1}^{1000}E[S_i]=\sum_{i=1}^{1000}E[S_i]$

where $S_i$ denotes expected marks of a Student i.

Now to calculate the expected marks of a student, we need to take Union over expectations of marks from each question

$E[S_i]=\bigcup_{j=1}^{150}E[Q_i]=\sum _{j=1}^{150}E[Q_i]$

where $E[Q_i]$ indicates expected marks per question.

Now with probabilty $\frac{1}{4}$ we can correctly mark a question and fetch +1 mark or with probability $\frac{3}{4}$ we can incorrectly mark a question and loose -0.25 marks.

So Expected marks per question=$\left ( \frac{1}{4}*1 \right )+\left ( \frac{3}{4} *-0.25\right )=0.0625=E[Q_i]$

So, $E[S_i]=150*0.0625=9.375$

And sum total of expected marks of all students=1000*9.375=9375.
reshown

@ANSR1010

It is given that all questions are MCQ .

Multiple choice - Wikipedia

Randomly means that the students are not biased towards any particular option.

Uniform Probability means that the probability of any particular option being chosen is the same. Which is 1/4 in this case.

It is a Binomial experiment as we are doing the same bernouli experiment repeatedly.

Which is the Bernouli experiment here ? Visiting a question and checking whether it is a correct answer or not. Why it is a Bernouli experiment ? Because a question can have either correct answer(success) or wrong answer(failure).

Total number of questions we are visiting,n = 1000*150 = 150000.

Let X be a random variable which indicates number of correct answer among the total of 150000 questions.

E(X) (i.e) Expectation of X (i.e) Expected number of correct answers = n * p  (since it is a binomial experiment)

where n  is no: of Bernouli experiments and p is probability of success (i.e) probability of correct answer.

Here n = 150000 , p = 0.25

So,

Expected number of correct answers = 150000 * .25 = 37500.

Expected number of wrong answers = 150000 - 37500 = 112500.

Expected Total marks = Expected number of correct answers * number of marks per correct answer + Expected number of wrong answers * number of marks per wrong answer.

= [37500 * 1] + [112500 * (-.25)]

= 37500 - 28125

= 9375

Expected marks per question is = -0.25 * 3/4 + 1 * 1/4 = 1/16
Since choice is uniformly distributed, expected marks = 150*1000/16 = 9375

Firstly, pivot one individual

Let X be the number of questions attempted correctly by this person

$E[X] = np = 150 \times \frac{1}{4} = 37.5$

Expected marks = $37.5 – (150-37.5)*0.25$ = 9.375

expected marks will be same for all 1000 students,

sum of expected marks will be 1000*9.375 = 9375

Number of question attempted correctly by each individual is a binomial’s distribution, (then marks is just a linear function over number of questions attempted, which then again will be binomial) and it is the combination of 150 Bernoulli’s distribution one for each individual question in the exam, (This is the idea behind ans given by Arjun Sir)

Let $X_i$ defines the marks scored by $i^{th}$  person let final result X, would be defined by

$\displaystyle X = \sum_{i=1}^{1000} X_i$

$\displaystyle E(X) = E( \sum_{i=1}^{1000} X_i) = 1000\cdot X_1$