PUNE MCA Exam 2016 Question Paper with answers for online practice

Concept:The probability that the winner of a game wins the next game is $\frac{2}{3}$, and the other wins with $\frac{1}{3}$. We need the chance that the first-game winner wins at least 3 of the next 4 games.Explanation:Let game 1 winner be A. For games 2–5, we want A to win 3 or 4 times.• All 4 wins: probability $\left(\frac{2}{3}\right)^4 = \frac{16}{81}$.• Exactly 3 wins: one loss occurs in game 2, 3, 4, or 5.• Loss in game 2: $\frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{81}$.• Loss in game 3: $\frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{4}{81}$.• Loss in game 4: $\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} = \frac{4}{81}$.• Loss in game 5: $\frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{1}{3} = \frac{8}{81}$. Sum for exactly 3 wins: $\frac{4+4+4+8}{81} = \frac{20}{81}$.• Total for at least 3 wins: $\frac{16}{81} + \frac{20}{81} = \frac{36}{81} = \frac{4}{9}$.Answer:$\frac{4}{9}$