GATE CS 2021 Set 2 Q.39 (CS Q.29): Probability and Statistics

In an examination, a student can choose the order in which two questions (QuesA and QuesB) must be attempted.

If the first question is answered wrong, the student gets zero marks.
If the first question is answered correctly and the second question is not answered correctly, the student gets the marks only for the first question.
If both the questions are answered correctly, the student gets the sum of the marks of the two questions.

The following table shows the probability of correctly answering a question and the marks of the question respectively.

Question	Probability of answering correctly	Marks
QuesA	0.8	10
QuesB	0.5	20

Assuming that the student always wants to maximize her expected marks in the examination, in which order should she attempt the questions and what is the expected marks for that order (assume that the questions are independent)?

First QuesA and then QuesB. Expected marks 14.

First QuesB and then QuesA. Expected marks 14.

First QuesB and then QuesA. Expected marks 22.

First QuesA and then QuesB. Expected marks 16.

Official answer and solution

Official answer: D
From: Official GATE 2021 answer key

Solution

Calculate the expected marks for both possible question attempt orders and choose the order that maximizes the expected score.

Let $A$ denote correctly answering QuesA ( $P(A) = 0.8$ , marks = 10), and $B$ denote correctly answering QuesB ( $P(B) = 0.5$ , marks = 20). Since $A$ and $B$ are independent:

$P(A) = 0.8, P(\overline{A}) = 0.2$
$P(B) = 0.5, P(\overline{B}) = 0.5$

Case 1: Attempt QuesA first, then QuesB:

If QuesA is wrong ( $P = 0.2$ ): marks = 0.
If QuesA is right and QuesB is wrong ( $P = 0.8 \times 0.5 = 0.4$ ): marks = 10.
If both QuesA and QuesB are right ( $P = 0.8 \times 0.5 = 0.4$ ): marks = $10 + 20 = 30$ .

E_1 = (0.2 \times 0) + (0.4 \times 10) + (0.4 \times 30) = 0 + 4 + 12 = 16

Case 2: Attempt QuesB first, then QuesA:

If QuesB is wrong ( $P = 0.5$ ): marks = 0.
If QuesB is right and QuesA is wrong ( $P = 0.5 \times 0.2 = 0.1$ ): marks = 20.
If both QuesB and QuesA are right ( $P = 0.5 \times 0.8 = 0.4$ ): marks = $20 + 10 = 30$ .

E_2 = (0.5 \times 0) + (0.1 \times 20) + (0.4 \times 30) = 0 + 2 + 12 = 14

Comparing the two expected values, $E_1 = 16 > E_2 = 14$ . The student should attempt QuesA first and then QuesB, giving expected marks of 16. This matches option D.

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2021 Computer Science and Information Technology - Session 2 (CS2)
Question: 39 of 65, printed as CS Q.29
Section: Computer Science and Information Technology

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