Probability in Game | AMC-10A, 2005 | Problem 18

Given that  The first team to win three games wins the series, team B wins the second game and team A wins the series. So the Total number of games played=\(5\). Now we have to find out the possible order of wins.....

Can you now finish the problem ..........

Possible cases :

If team B won the first two games, team A would need to win the next three games. Therefore the possible order of wins is BBAAA.
If team A won the first game, and team B won the second game, the possible order of wins is $A B B A A, A B A B A,$ and $A B A A X,$ where $X$ denotes that the 5th game wasn't played.
since ABAAX is dependent on the outcome of 4 games instead of 5, it is twice as likely to occur and can be treated as two possibilities.

According to the question, there is One possibility where team $\mathrm{B}$ wins the first game and 5 total possibilities, Therefore the required probability is \(\frac{ 1}{5}\)