Difference between revisions of "2016 AMC 12B Problems/Problem 19"
(→Solution) |
Pi is 3.14 (talk | contribs) (→Solution 2) |
||
(12 intermediate revisions by 5 users not shown) | |||
Line 9: | Line 9: | ||
\textbf{(E)}\ \frac{1}{3}</math> | \textbf{(E)}\ \frac{1}{3}</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | + | By: dragonfly | |
− | <math> | + | 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 |
− | And then we find that <math>x</math> equals to <math>\boxed{\textbf{(B)}\ \frac{1}{7}}</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> because of the formula of the sum for an infinite series, <math>\frac{\frac{1}{8}}{1-\frac{1}{8}} = \frac{1}{8}*\frac{8}{7} = \frac{1}{7}</math>. | ||
+ | |||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Call it a "win" if the boys all flip their coins the same number of times, and the probability that they win is <math>P</math>. The probability that they win on their first flip is <math>\frac{1}{8}</math>. If they don't win on their first flip, that means they all flipped tails (which also happens with probability <math>\frac{1}{8}</math> ) and that their chances of winning have returned to what they were at the beginning. This covers all possible sequences of winning flips. So we have | ||
+ | |||
+ | <math>P = \frac{1}{8} + \frac{1}{8}P</math> | ||
+ | |||
+ | Solving for <math>P</math> gives <math>\boxed{\textbf{(B)}\ \frac{1}{7}}</math>. | ||
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/rLAcJe3o-uA?t=75 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2016|ab=B|num-b=18|num-a=20}} | {{AMC12 box|year=2016|ab=B|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 07:27, 4 November 2022
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 1
By: dragonfly
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
because of the formula of the sum for an infinite series,
.
Solution 2
Call it a "win" if the boys all flip their coins the same number of times, and the probability that they win is . The probability that they win on their first flip is
. If they don't win on their first flip, that means they all flipped tails (which also happens with probability
) and that their chances of winning have returned to what they were at the beginning. This covers all possible sequences of winning flips. So we have
Solving for gives
.
Video Solution by OmegaLearn
https://youtu.be/rLAcJe3o-uA?t=75
~ pi_is_3.14
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.