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

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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.  
 
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 <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 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>.
  
 
==Note==
 
==Note==
−
The problem is poorly worded, since the problem directly states that the answer is <math>\boxed{1/2}</math>.
+
The original final problem was poorly worded, since the problem directly stated 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?"
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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?"
 +
 
 +
(Many problems in probability are poorly worded.)
  
 
==See Also==
 
==See Also==

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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