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

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