Try this beautiful problem from AMC 10A, 2005 based on Probability in Game.
Probability in Game – AMC-10A, 2005- Problem 18
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?
- \(\frac{1}{4}\)
- \(\frac{1}{6}\)
- \(\frac{1}{5}\)
- \(\frac{2}{3}\)
- \(\frac{1}{3}\)
Key Concepts
Probability
combinatorics
Check the Answer
But try the problem first…
Answer: \(\frac{1}{5}\)
AMC-10A (2005) Problem 18
Pre College Mathematics
Try with Hints
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 \(\frac{ 1}{5}\)
Other useful links
- 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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