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Try this beautiful problem from AMC 10A, 2005 based on Probability in Game.

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?

- \(\frac{1}{4}\)
- \(\frac{1}{6}\)
- \(\frac{1}{5}\)
- \(\frac{2}{3}\)
- \(\frac{1}{3}\)

Probability

combinatorics

But try the problem first...

Answer: \(\frac{1}{5}\)

Source

Suggested Reading

AMC-10A (2005) Problem 18

Pre College Mathematics

First hint

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 ..........

Second Hint

**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.

Final Step

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 \(\frac{ 1}{5}\)

- https://www.cheenta.com/pentagon-square-pattern-amc-10a-2001-problem-18/
- https://www.youtube.com/watch?v=U_LztQXd12A&t=119s

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