COMEDK 2025 Evening Solved Paper

Concept:The expected value is the sum of probabilities of each outcome multiplied by the corresponding gain or loss.Explanation:Let $p$ be the probability of winning on a single throw: $p = \frac{2}{6} = \frac{1}{3}$ (even numbers $\ge 4$ are 4 and 6).Let $q = 1 - p = \frac{2}{3}$ be the probability of losing on a throw.A win gives $+₹1000$; each loss gives $-\text{Rs. }200$.The player stops after a win or after 3 throws.Possible outcomes and their gains:• Win on 1st throw: probability $p$, gain $+1000$.• Lose 1st, win 2nd: probability $q \cdot p$, gain $-200 + 1000 = +800$.• Lose 1st & 2nd, win 3rd: probability $q^2 \cdot p$, gain $-400 + 1000 = +600$.• Lose all three: probability $q^3$, gain $-600$.Expected gain $E[G]$ is:$$E[G] = p \cdot 1000 + (q p) \cdot 800 + (q^2 p) \cdot 600 + q^3 \cdot (-600).$$Substitute $p = \frac{1}{3}$, $q = \frac{2}{3}$:$$E[G] = \frac{1}{3} \cdot 1000 + \left(\frac{2}{3} \cdot \frac{1}{3}\right) \cdot 800 + \left(\left(\frac{2}{3}\right)^2 \cdot \frac{1}{3}\right) \cdot 600 + \left(\frac{2}{3}\right)^3 \cdot (-600).$$Simplify each term:$\frac{1000}{3} = \frac{3000}{9}$,$\frac{2}{9} \cdot 800 = \frac{1600}{9}$,$\frac{4}{27} \cdot 600 = \frac{2400}{27} = \frac{800}{9}$,$\frac{8}{27} \cdot (-600) = -\frac{4800}{27} = -\frac{1600}{9}$.Sum: $\frac{3000 + 1600 + 800 - 1600}{9} = \frac{3800}{9}$.Thus the expected gain is $\frac{3800}{9}$ rupees.Answer:Option B: $\frac{3800}{9}$ gain.