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

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==Solution 2 (99% accurate guesswork)==
 
==Solution 2 (99% accurate guesswork)==
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Let's try the answer choices. We can quickly find that when we roll <math>3</math> dice, either the first and second sum to <math>7</math>, the first and third sum to <math>7</math>, or the second and third sum to <math>7</math>. There are <math>6</math> ways for the first and second dice to sum to <math>7</math>,  <math>6</math> ways for the first and third to sum to <math>7</math>, and <math>6</math> ways for the second and third dice to sum to <math>7</math>. However, we overcounted (but not by much) so we can assume that the answer is <math>\boxed {\textbf{(C) 7}}</math>
+
Let's try the answer choices. We can quickly find that when we roll <math>3</math> dice, either the first and second sum to <math>7</math>, the first and third sum to <math>7</math>, or the second and third sum to <math>7</math>. There are <math>6</math> ways for the first and second dice to sum to <math>7</math>,  <math>6</math> ways for the first and third to sum to <math>7</math>, and <math>6</math> ways for the second and third dice to sum to <math>7</math>. However, we overcounted (but not by much) so we can assume that the answer is <math>\boxed {\textbf{(C) 4}}</math>
  
 
~Arcticturn
 
~Arcticturn

Revision as of 07:24, 19 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 $\left(\frac56\right)^4<\frac12<\left(\frac56\right)^3,$ 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

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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