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Difference between revisions of "2005 AMC 10A Problems/Problem 18"
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There is <math>1</math> possibility where team <math>B</math> wins the first game and <math>4</math> total possibilities when team <math>A</math> wins the series and team <math>B</math> wins the second game. Note that the fourth possibility <math>(ABAAX)</math> occurs twice as often as the others, so we put <math>1</math> over <math>5</math> total possibilities. The desired probability is then <math>\boxed{\textbf{(A) }\frac{1}{5}}</math>.
 
There is <math>1</math> possibility where team <math>B</math> wins the first game and <math>4</math> total possibilities when team <math>A</math> wins the series and team <math>B</math> wins the second game. Note that the fourth possibility <math>(ABAAX)</math> occurs twice as often as the others, so we put <math>1</math> over <math>5</math> total possibilities. The desired probability is then <math>\boxed{\textbf{(A) }\frac{1}{5}}</math>.
  
 +
==Note==
 +
The problem is poorly worded, since the problem directly states that the answer is <math>\boxed{1/2}</math>.
  
 +
The problem should say "what fraction of possible sets of game outcomes have <math>B</math> winning the first game?" or "What is the *posterior* probability that <math>B</math> won the first game?"
  
 
==See Also==
 
==See Also==

Revision as of 23:50, 8 July 2024

Problem

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?

$\textbf{(A) } \frac{1}{5}\qquad \textbf{(B) } \frac{1}{4}\qquad \textbf{(C) } \frac{1}{3}\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \frac{2}{3}$

Solution

There are at most $5$ games played.

If team $B$ won the first two games, team $A$ would need to win the next three games. So the only 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 are: $ABBAA, ABABA,$ and $ABAAX$, where $X$ denotes that the $5$th game wasn't played.

There is $1$ possibility where team $B$ wins the first game and $4$ total possibilities when team $A$ wins the series and team $B$ wins the second game. Note that the fourth possibility $(ABAAX)$ occurs twice as often as the others, so we put $1$ over $5$ total possibilities. The desired probability is then $\boxed{\textbf{(A) }\frac{1}{5}}$.

Note

The problem is poorly worded, since the problem directly states that the answer is $\boxed{1/2}$.

The problem should say "what fraction of possible sets of game outcomes have $B$ winning the first game?" or "What is the *posterior* probability that $B$ won the first game?"

See Also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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All AMC 10 Problems and Solutions

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