Question

Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?

$\dfrac{1}{5}$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

Answer: A

Concepts:conditional probability systematic listing

Difficulty rating: 1860

Solution:

Suppose all five games are played, so every sequence of five results is equally likely. Requiring that B wins game $2$ and A ends up with the series (three wins) leaves the equally likely sequences

$\begin{gathered} \text{BBAAA}, \quad \text{ABBAA}, \\ \text{ABABA}, \quad \text{ABAAB}, \\ \text{ABAAA}. \end{gathered}$

Only in BBAAA does team B win the first game, so the probability is $\dfrac{1}{5}.$

Thus, the correct answer is A.

2005 AMC 10A Problem 18

Answer: A

Difficulty rating: 1860

Solution:

Problem 18 in Other Years