Art of Problem Solving
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
During AMC 10A/12A testing, the AoPS Wiki is in read-only mode and no edits can be made.

2022 AMC 10B Problems/Problem 12: Difference between revisions

← Older editNewer edit →
MRENTHUSIASM (talk | contribs)
editor
9,986 edits
No edit summary
MRENTHUSIASM (talk | contribs)
editor
9,986 edits
No edit summary
Line 5: Line 5:


==Solution==
==Solution==
Rolling a pair of fair <math>6</math>-sided dice, the probability of getting a sum of <math>7</math> is <math>\frac16:</math> Regardless what the first die shows, the second die has exactly one possibility to make the sum <math>7.</math>
Rolling a pair of fair <math>6</math>-sided dice, the probability of getting a sum of <math>7</math> is <math>\frac16:</math> Regardless what the first die shows, the second die has exactly one outcome to make the sum <math>7.</math> We consider the complement: The probability of not getting a sum of <math>7</math> is <math>1-\frac16=\frac56.</math> Rolling the pair of dice <math>n</math> times, the probability of getting a sum of <math>7</math> at least once is <math>1-\left(\frac56\right)^n.</math>
 
Therefore, we have <math>1-\left(\frac56\right)^n>\frac12,</math> or <cmath>\left(\frac56\right)^n<\frac12.</cmath> Since, <math>\left(\frac56\right)^3>\frac12>\left(\frac56\right)^4,</math> the least integer <math>n</math> satisfying the inequality is <math>\boxed{\textbf{(C) } 4}.</math>


~MRENTHUSIASM
~MRENTHUSIASM

Revision as of 18:14, 17 November 2022

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

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, $\left(\frac56\right)^3>\frac12>\left(\frac56\right)^4,$ the least integer $n$ satisfying the inequality is $\boxed{\textbf{(C) } 4}.$

~MRENTHUSIASM

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_12&oldid=182028"