Difference between revisions of "2016 AMC 12B Problems/Problem 19"
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We can solve this problem by listing it as an infinite geometric equation. We get that to have the same amount of tosses, they have a <math>\frac{1}{8}</math> chance of getting all heads. Then the next probability is all of them getting tails and then on the second try, they all get heads. The probability of that happening is <math>\left(\frac{1}{8}\right)^2</math>.We then get the geometric equation | We can solve this problem by listing it as an infinite geometric equation. We get that to have the same amount of tosses, they have a <math>\frac{1}{8}</math> chance of getting all heads. Then the next probability is all of them getting tails and then on the second try, they all get heads. The probability of that happening is <math>\left(\frac{1}{8}\right)^2</math>.We then get the geometric equation | ||
| − | <math>x=\frac{1}{8}+(\frac{1}{8})^2+(\frac{1}{8})^3...</math> | + | <math>x=\frac{1}{8}+\left(\frac{1}{8}\right)^2+\left(\frac{1}{8}\right)^3...</math> |
And then we find that <math>x</math> equals to <math>\boxed{\textbf{(B)}\ \frac{1}{7}}</math> | And then we find that <math>x</math> equals to <math>\boxed{\textbf{(B)}\ \frac{1}{7}}</math> | ||
Revision as of 16:05, 21 February 2016
Problem
Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips a fair coin repeatedly until he gets his first head, at which point he stops. What is the probability that all three flip their coins the same number of times?
Solution
We can solve this problem by listing it as an infinite geometric equation. We get that to have the same amount of tosses, they have a 
chance of getting all heads. Then the next probability is all of them getting tails and then on the second try, they all get heads. The probability of that happening is 
.We then get the geometric equation
And then we find that 
equals to 
See Also
| 2016 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 18 |
Followed by Problem 20 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
