GATE CS 2021 Set 2 Q29 — Probability: Random Variables and Distributions

To find the optimal strategy, we calculate the expected marks for both possible orders of attempting the questions.

Let the probabilities and marks for each question be:

• For $\textsf{QuesA}$: $P(A) = 0.8$, Marks $M_A = 10$
• For $\textsf{QuesB}$: $P(B) = 0.5$, Marks $M_B = 20$

Case 1: Attempting $\textsf{QuesA}$ first, then $\textsf{QuesB}$

The possible outcomes and their corresponding probabilities and marks are:

1. First question ($\textsf{QuesA}$) is wrong:
   Probability: $1 - P(A) = 1 - 0.8 = 0.2$
   Marks: $0$
2. $\textsf{QuesA}$ is correct, but $\textsf{QuesB}$ is wrong:
   Probability: $P(A) \times (1 - P(B)) = 0.8 \times 0.5 = 0.4$
   Marks: $M_A = 10$
3. Both questions are correct:
   Probability: $P(A) \times P(B) = 0.8 \times 0.5 = 0.4$
   Marks: $M_A + M_B = 10 + 20 = 30$

The expected marks for this order ($E_{AB}$) is:

Case 2: Attempting $\textsf{QuesB}$ first, then $\textsf{QuesA}$

The possible outcomes and their corresponding probabilities and marks are:

1. First question ($\textsf{QuesB}$) is wrong:
   Probability: $1 - P(B) = 1 - 0.5 = 0.5$
   Marks: $0$
2. $\textsf{QuesB}$ is correct, but $\textsf{QuesA}$ is wrong:
   Probability: $P(B) \times (1 - P(A)) = 0.5 \times 0.2 = 0.1$
   Marks: $M_B = 20$
3. Both questions are correct:
   Probability: $P(B) \times P(A) = 0.5 \times 0.8 = 0.4$
   Marks: $M_B + M_A = 20 + 10 = 30$

The expected marks for this order ($E_{BA}$) is:

Conclusion

Comparing the expected marks:

• Order $\textsf{QuesA} \rightarrow \textsf{QuesB}$ yields an expected score of $16$.
• Order $\textsf{QuesB} \rightarrow \textsf{QuesA}$ yields an expected score of $14$.

To maximize her expected marks, the student should attempt First $\textsf{QuesA}$ and then $\textsf{QuesB}$, resulting in an expected score of $16$.