Difference between revisions of "1993 AIME Problems/Problem 11"
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Since <math>a_6=\frac{364}{729}</math>, <math>m+n = 1093 \equiv \boxed{093} \pmod{1000}</math>. | Since <math>a_6=\frac{364}{729}</math>, <math>m+n = 1093 \equiv \boxed{093} \pmod{1000}</math>. | ||
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+ | ==Solution 2== | ||
== See also == | == See also == | ||
{{AIME box|year=1993|num-b=10|num-a=12}} | {{AIME box|year=1993|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 16:37, 25 February 2018
Contents
Problem
Alfred and Bonnie play a game in which they take turns tossing a fair coin. The winner of a game is the first person to obtain a head. Alfred and Bonnie play this game several times with the stipulation that the loser of a game goes first in the next game. Suppose that Alfred goes first in the first game, and that the probability that he wins the sixth game is , where and are relatively prime positive integers. What are the last three digits of ?
Solution
The probability that the th flip in each game occurs and is a head is . The first person wins if the coin lands heads on an odd numbered flip. So, the probability of the first person winning the game is , and the probability of the second person winning is .
Let be the probability that Alfred wins the th game, and let be the probability that Bonnie wins the th game.
If Alfred wins the th game, then the probability that Alfred wins the th game is . If Bonnie wins the th game, then the probability that Alfred wins the th game is .
Thus, .
Similarly, .
Since Alfred goes first in the st game, .
Using these recursive equations:
Since , .
Solution 2
See also
1993 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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