GATE CSE 2004 | Question: 74

Question

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

• A. $0$
• B. $2550$
• C. $7525$
• D. $9375$

Comments

Real-life application.

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Ans

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With negative marking of –0.33 per question (GATE style), the expected score magically becomes 0.
Clearly, GATE folks have done their math very well 😄

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@VivekVadadoriya Brother,

What a vision, what a thought 😂

But I think, for 1 mark questions with -0.33 negative marking,
E = (0.25 ×1)+(0.75×-0.33).
E = 0.25 - 0.2475.
E =0.0025.

Which means gate exam apearing students, after doing all the mental gymnastics and stuff with the question paper, our average expected gain in marks will be 0.0025 😂

To any GATE aspirant out there, this is purely for fun and meant to bring a smile 😊. There is no intention to offend anyone.

Answer 1

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 = $9.375 \times 1000 = 9375$.

Comments

@Rakesh K 

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.

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That is exactly why we have -0.33 marks for a wrong attempt

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Best answer indeed.

Answer 2

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. 

Answer 3

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

Answer 4

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

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Just explaining few more facts about this problems,

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 E(X_1) $

Comments

Does the approach changes if

"1000 students choose all their answers randomly without uniform probability"

??

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What do you mean by without uniform probability?

If you mean that the probability of choosing each option is not uniform, then no, the approach will not change given that for each question, the probability of choosing the correct option is the same, and he solves them independently;

just put the probability of selecting the correct option in place of $\frac{1}{4}$.

FYI if question says non-uniform probability then it should provide a way to find the probability of choosing the right option for each question. Which just saying without uniform probabilty does not gives.

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thank u , Understood👍

Answer 5

Ans

Answer 6

Expected marks = E[X]
E[X] = E[X1] + E[X2] + E[X3] + ... E[X150] where E[Xi] is expected marks of question i

E[X1] = 0.25 * 1 + 0.75 * -0.25 = 0.0625 = E[X2]  = E[X3] = ..... = E[X150]

E[X] = 150 * 0.0625 = 9.375
Total sum of expected marks for 1000 students = 1000*9.375 = 9375