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Difference between revisions of "1971 AHSME Problems/Problem 23"

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== Solution ==
 
== Solution ==
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<math>\boxed{\textbf{(A) } 11 \text{ to } 5}</math>.
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We have two cases: one where <math>A</math> wins the first game and the other where <math>A</math> loses.
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<math>\underline{\text{Case 1:}}</math> In the <math>\tfrac12</math> chance that <math>A</math> wins the first game, <math>A</math> simply needs to win at least <math>1</math> of the next <math>3</math> games. We see that the probability of <math>A</math> losing the next <math>3</math> games is <math>(\tfrac12)^3=\tfrac18</math>, so, by [[complementary counting]], the probability that <math>A</math> wins at least <math>1</math> of the next <math>3</math> games is <math>1-\tfrac18=\tfrac78</math>.
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<math>\underline{\text{Case 2:}}</math> In the <math>\tfrac12</math> chance that <math>A</math> loses the first game, both teams need to win <math>2</math> games, so <math>A</math>'s advantage completely disappears. Thus, the proability that <math>A</math> wins the series from here is <math>\tfrac12</math>.
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Combining the information from the two cases, we see that <math>A</math>'s probability of winning the series is <math>\tfrac12 \cdot \tfrac78 + \tfrac12 \cdot \tfrac12 = \tfrac7{16} + \tfrac4{16} = \tfrac{11}{16}</math>. Thus, our answer is <math>\boxed{\textbf{(A) } 11 \text{ to } 5}</math>.
  
 
== See Also ==
 
== See Also ==
 
{{AHSME 35p box|year=1971|num-b=22|num-a=24}}
 
{{AHSME 35p box|year=1971|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 18:36, 6 August 2024

Problem

Teams $A$ and $B$ are playing a series of games. If the odds for either to win any game are even and Team $A$ must win two or Team $B$ three games to win the series, then the odds favoring Team $A$ to win the series are

$\textbf{(A) }11\text{ to }5\qquad \textbf{(B) }5\text{ to }2\qquad \textbf{(C) }8\text{ to }3\qquad \textbf{(D) }3\text{ to }2\qquad \textbf{(E) }13\text{ to }6$

Solution

We have two cases: one where $A$ wins the first game and the other where $A$ loses.

$\underline{\text{Case 1:}}$ In the $\tfrac12$ chance that $A$ wins the first game, $A$ simply needs to win at least $1$ of the next $3$ games. We see that the probability of $A$ losing the next $3$ games is $(\tfrac12)^3=\tfrac18$, so, by complementary counting, the probability that $A$ wins at least $1$ of the next $3$ games is $1-\tfrac18=\tfrac78$.

$\underline{\text{Case 2:}}$ In the $\tfrac12$ chance that $A$ loses the first game, both teams need to win $2$ games, so $A$'s advantage completely disappears. Thus, the proability that $A$ wins the series from here is $\tfrac12$.

Combining the information from the two cases, we see that $A$'s probability of winning the series is $\tfrac12 \cdot \tfrac78 + \tfrac12 \cdot \tfrac12 = \tfrac7{16} + \tfrac4{16} = \tfrac{11}{16}$. Thus, our answer is $\boxed{\textbf{(A) } 11 \text{ to } 5}$.

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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