Difference between revisions of "1971 AHSME Problems/Problem 23"
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== Solution == | == Solution == | ||
− | <math>\boxed{\textbf{(A) } 11 \text{ to } 5}</math>. | + | 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>. | ||
+ | |||
+ | <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 and
are playing a series of games. If the odds for either to win any game are even and
Team
must win two or Team
three games to win the series, then the odds favoring Team
to win the series are
Solution
We have two cases: one where wins the first game and the other where
loses.
In the
chance that
wins the first game,
simply needs to win at least
of the next
games. We see that the probability of
losing the next
games is
, so, by complementary counting, the probability that
wins at least
of the next
games is
.
In the
chance that
loses the first game, both teams need to win
games, so
's advantage completely disappears. Thus, the proability that
wins the series from here is
.
Combining the information from the two cases, we see that 's probability of winning the series is
. Thus, our answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.