A random attempt at JEE Mains!

Probability Level 4

In an examination, a student is faced with five alternatives. He can either mark options A, B, C, D or leave the question unattempted. Each of the five alternatives are assumed to be equally likely!! There are 90 questions.

There is only one correct option out A, B, C, and D.

A correct response is rewarded +4 points/question; a wrong response, -1 points/question; an unattempted question yields 0 points.

Given that the student randomly attempts the exam, what is his expected score?

The answer is 18.

2 solutions

M M
May 6, 2017

Expected outcome of one trial:

Points	Probability	$x_i p_i$
4	$\frac{1}{5}$	$\frac{4}{5}$
0	$\frac{1}{5}$	0
-1	$\frac{3}{5}$	- $\frac{3}{5}$

Expected value from one question: $\frac{4}{5} - \frac{3}{5} = \frac{1}{5}$

Since expected value is linear , taking a sum by sampling from this distribution 90 times has expected value $90 \cdot 1/5 = 18$

Chew-Seong Cheong
Apr 29, 2017

The probability that an option is marked or not marked is $p_a =p_b = p_c = p_d = p_0 = \frac 15$ .
The probability that what marked is correct $q_c = \frac 14$ .
The probability that what marked is incorrect $q_w = \frac 34$ .
Therefore the expected score for a question $E_1 = \left(p_a+p_b+p_c+p_d\right)\left(4q_c - 1q_w\right) + p_0(0)$
And the expected score for all 90 question:

$\begin{aligned} \quad \implies E_{90} & = 90\left(\left(p_a+p_b+p_c+p_d\right)\left(4q_c - 1q_w\right) + p_0(0)\right) \\ & = 90\left(\left(\frac 45 \right) \left(4 \cdot \frac 14 - 1 \cdot \frac 34 \right) + 0 \right) \\ & = \boxed{18} \end{aligned}$

This is definitely more elegant than what I did.

$\large \sum_{r=0}^{90} \sum_{m=0}^{90-r} ^{90}C_{r}^{90-r}C_{m} (\frac{3}{5})^{m}(\frac{1}{5})^{90-m}(4r-m)$

Rohith M.Athreya - 4 years, 1 month ago

A random attempt at JEE Mains!

The answer is 18.

2 solutions

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