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Difference between revisions of "2005 AMC 10A Problems/Problem 18"

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==Problem==
 
==Problem==
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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?  
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Team <math>A</math> and team <math>B</math> 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 <math>B</math> wins the second game and team <math>A</math> wins the series, what is the probability that team <math>B</math> wins the first game?  
  
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<math> \mathrm{(A) \ } \frac{1}{5}\qquad \mathrm{(B) \ } \frac{1}{4}\qquad \mathrm{(C) \ } \frac{1}{3}\qquad \mathrm{(D) \ } \frac{1}{2}\qquad \mathrm{(E) \ } \frac{2}{3} </math>
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<math>
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\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}
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</math>
  
 
==Solution==
 
==Solution==
 
There are at most <math>5</math> games played.  
 
There are at most <math>5</math> games played.  
  
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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.  
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If team <math>B</math> won the first two games, team <math>A</math> would need to win the next three games. So the only possible order of wins is <math>BBAAA</math>.  
  
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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 5th game wasn't played.  
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If team <math>A</math> won the first game, and team <math>B</math> won the second game, the possible orders of wins are <math>ABBAA</math>, <math>ABABA</math>, and <math>ABAAX</math>, where <math>X</math> denotes that the <math>5</math>th game wasn't played.  
  
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Since ABAAX is dependent on the outcome of <math>4</math> games instead of <math>5</math>, it is twice as likely to occur and can be treated as two possibilities. (If <math>5 </math> games were played, X could be A or B)
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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 because it is dependent on the outcome of only <math>4</math> games rather than <math>5</math>, so we put <math>1</math> over <math>1 \cdot 2 + (4-1) = 5</math> total possibilities. The desired probability is then <math>\boxed{\textbf{(A) } \frac{1}{5}}</math>.
  
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Since there is <math>1</math> possibility where team B wins the first game and <math>5</math> total possibilities, the desired probability is <math>\frac{1}{5}\Rightarrow \boxed{A}</math>
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==Note==
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The original final problem was poorly worded, since the problem directly stated that the answer is <math>\boxed{1/2}</math> (the probability of team <math>B</math> simply winning the first game, as opposed to the desired conditional probability of this event ''given that'' both team <math>B</math> wins the second game and team <math>A</math> wins the series).
  
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CHECK OUT Video Solution: https://youtu.be/xqdc2N6fckA
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The problem should therefore say something like "What fraction of possible sets of game outcomes have <math>B</math> winning the first game?" or "Given the observed results, what is the conditional probability that <math>B</math> won the first game?"
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(Many problems in probability are poorly worded.)
  
 
==See Also==
 
==See Also==
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{{AMC10 box|year=2005|ab=A|num-b=17|num-a=19}}
 
{{AMC10 box|year=2005|ab=A|num-b=17|num-a=19}}
  
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[[Category:Introductory Geometry Problems]]
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[[Category:Introductory Combinatorics Problems]]
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[[Category:Area Ratio Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 01:13, 29 July 2025

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 orders 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 because it is dependent on the outcome of only $4$ games rather than $5$, so we put $1$ over $1 \cdot 2 + (4-1) = 5$ total possibilities. The desired probability is then $\boxed{\textbf{(A) } \frac{1}{5}}$.

Note

The original final problem was poorly worded, since the problem directly stated that the answer is $\boxed{1/2}$ (the probability of team $B$ simply winning the first game, as opposed to the desired conditional probability of this event given that both team $B$ wins the second game and team $A$ wins the series).

The problem should therefore say something like "What fraction of possible sets of game outcomes have $B$ winning the first game?" or "Given the observed results, what is the conditional probability that $B$ won the first game?"

(Many problems in probability are poorly worded.)

See Also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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