IIT-JEE 1989Question 1
Suppose the probability for $A$ to win a game against $B$ is $0.4$. If $A$ has an option of playing either a 'best of $3$ games' or a 'best of $5$ games' match against $B$, which option should he choose so that the probability of his winning the match is higher? (No game ends in a draw).
Probability
AnswerBest of $3$ games (probability $0.352$ versus $0.31744$ for best of $5$)
Solution
Best of 3: $3(0.4)^2(0.6)+(0.4)^3=0.352$. Best of 5: $\sum_{i\ge3}\binom5i(0.4)^i(0.6)^{5-i}=0.31744$. So he should choose the best of $3$ match.