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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 A and team B play a series. The first team to win three games wins the series. Before each game, each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If it turns out that team B won the second game and team A won the series, what is the conditional probability that team B won 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> \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} </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 order of wins are: <math>ABBAA, 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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There is <math>1</math> possibility where team B wins the first game and <math>4</math> total possibilities when team A wins the tournament and team B wins the second game. However, note that the fourth possibility (ABAAX) occurs twice as often as the others, so we put <math>1</math> over <math>5</math> total possibilities instead of <math>4</math>. the desired probability is then <math>\frac{1}{5}\Rightarrow \boxed{A}.</math>
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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 <math>4</math> games instead of <math>5</math>, 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>.
  
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Note:  The solution given to this problem above is incorrect (you can check the answer key of this test to be sure of this). In actuality, the fourth possibility (ABAAX) simply counts as 1 case, and it does not occur twice as often as the others because even though we don't have to multiply by <math>1/2</math> at the end, we are given the information that team A will win by the information in the problem. Thus, we only have 4 cases, out of which only 1 (BBAAA) is desired. Thus, our answer is <math>\frac{1}{4}\Rightarrow \boxed{B}.</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>.
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The problem should say "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]]
+
[[Category:Introductory Combinatorics Problems]]
−
[[Category:Area Ratio Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 09:34, 23 July 2024

Problem

Team A and team B play a series. The first team to win three games wins the series. Before each game, each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If it turns out that team B won the second game and team A won the series, what is the conditional probability that team B won 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 because it is dependent on the outcome of $4$ games instead of $5$, so we put $1$ over $5$ 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 problem should say "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

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