Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 10B Problems/Problem 12"

(→‎Solution 1)
(→‎Video Solution 1)
Line 20: Line 20:
  
 
~Education, the Study of Everything
 
~Education, the Study of Everything
 +
==Video Solution by Interstigation==
 +
https://youtu.be/qT0hVzy7zeY
  
 
== See Also ==
 
== See Also ==
 
{{AMC10 box|year=2022|ab=B|num-b=11|num-a=13}}
 
{{AMC10 box|year=2022|ab=B|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 01:14, 28 April 2023

Problem

A pair of fair $6$-sided dice is rolled $n$ times. What is the least value of $n$ such that the probability that the sum of the numbers face up on a roll equals $7$ at least once is greater than $\frac{1}{2}$?

$\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6$

Solution 1 (Complement)

Rolling a pair of fair $6$-sided dice, the probability of getting a sum of $7$ is $\frac16:$ Regardless what the first die shows, the second die has exactly one outcome to make the sum $7.$ We consider the complement: The probability of not getting a sum of $7$ is $1-\frac16=\frac56.$ Rolling the pair of dice $n$ times, the probability of getting a sum of $7$ at least once is $1-\left(\frac56\right)^n.$

Therefore, we have $1-\left(\frac56\right)^n>\frac12,$ or \[\left(\frac56\right)^n<\frac12.\] Since $\left(\frac56\right)^4<\frac12<\left(\frac56\right)^3,$ the least integer $n$ satisfying the inequality is $\boxed{\textbf{(C) } 4}.$

~MRENTHUSIASM

Solution 2 (99% Accurate Guesswork)

Let's try the answer choices. We can quickly find that when we roll $3$ dice, either the first and second sum to $7$, the first and third sum to $7$, or the second and third sum to $7$. There are $6$ ways for the first and second dice to sum to $7$, $6$ ways for the first and third to sum to $7$, and $6$ ways for the second and third dice to sum to $7$. However, we overcounted (but not by much) so we can assume that the answer is $\boxed {\textbf{(C) }4}$

~Arcticturn

Video Solution 1

https://youtu.be/rYyb3NCWBXk

~Education, the Study of Everything

Video Solution by Interstigation

https://youtu.be/qT0hVzy7zeY

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_12&oldid=192572"