Difference between revisions of "2021 Fall AMC 10B Problems/Problem 20"
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| + | ==Problem 20== | ||
| + | In a particular game, each of <math>4</math> players rolls a standard <math>6{ }</math>-sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo's first roll was a <math>5,</math> given that he won the game? | ||
| + | |||
| + | <math>(\textbf{A})\: \frac{61}{216}\qquad(\textbf{B}) \: \frac{367}{1296}\qquad(\textbf{C}) \: \frac{41}{144}\qquad(\textbf{D}) \: \frac{185}{648}\qquad(\textbf{E}) \: \frac{11}{36}</math> | ||
| + | |||
| + | ==Solution 1== | ||
| + | |||
| + | Since we know that Hugo wins, we know that he rolled the highest number in the first round. The probability that his first roll is a <math>5</math> is just the probability that the highest roll in the first round is <math>5</math>. | ||
| + | |||
| + | Let <math>P(x)</math> indicate the probability that event <math>x</math> occurs. | ||
| + | We find that that <math>P(\text{No one rolls a 6})-P(\text{No one rolls a 5 or 6})=P(\text{The highest roll is a 5})</math>, | ||
| + | |||
| + | so <cmath>P(\text{No one rolls a 6})=\left(\frac56\right)^4,</cmath> | ||
| + | <cmath>P(\text{No one rolls a 5 or 6})=\left(\frac23\right)^4,</cmath> | ||
| + | <cmath>P(\text{The highest roll is a 5})=\left(\frac56\right)^4-\left(\frac46\right)^4=\frac{5^4-4^4}{6^4}=\frac{369}{1296}=\boxed{(\textbf{C}) \: \frac{41}{144}}.</cmath> | ||
| + | |||
| + | ~kingofpineapplz | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=21|num-b=19}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=21|num-b=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:46, 23 November 2021
Problem 20
In a particular game, each of 
players rolls a standard 
-sided die. The winner is the player who rolls the highest number. If there is a tie for the highest roll, those involved in the tie will roll again and this process will continue until one player wins. Hugo is one of the players in this game. What is the probability that Hugo's first roll was a 
given that he won the game?
Solution 1
Since we know that Hugo wins, we know that he rolled the highest number in the first round. The probability that his first roll is a 
is just the probability that the highest roll in the first round is .
Let 
indicate the probability that event 
occurs.
We find that that 
,
so ![\[P(\text{No one rolls a 6})=\left(\frac56\right)^4,\]](http://latex.artofproblemsolving.com/2/0/8/208b4a2de49251e6e306d6ce37bd168be78f1be9.png)
![\[P(\text{No one rolls a 5 or 6})=\left(\frac23\right)^4,\]](http://latex.artofproblemsolving.com/f/a/7/fa7be13fc7b1578b8eef2274364313c8228bb7cc.png)
![\[P(\text{The highest roll is a 5})=\left(\frac56\right)^4-\left(\frac46\right)^4=\frac{5^4-4^4}{6^4}=\frac{369}{1296}=\boxed{(\textbf{C}) \: \frac{41}{144}}.\]](http://latex.artofproblemsolving.com/8/9/f/89fb0bfcb40e4513d32aecf43a7cb77d4aec5e8c.png)
~kingofpineapplz
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
