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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) }4}</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
  
 +
==Solution 3==
 +
We can start by figuring out what the probability is for each die to add up to <math>7</math> if there is only <math>1</math> roll. We can quickly see that the probability is <math>\frac16</math>, as there are <math>6</math> ways to make <math>7</math> from <math>2</math> numbers on a die, and there are a total of <math>36</math> ways to add <math>2</math> numbers on a die. And since each time we roll the dice, we are adding to the probability, we can conclude that the total probability for rolling a sum of <math>7</math> in <math>n</math> rolls would be <math>\frac16</math><math>n</math>. The smallest number that satisfies this is <math>\boxed {\textbf{(C) }4}</math>.
 +
 +
~mihikamishra
  
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==Solution 3==
+
==Solution 4 (Fakesolve)==
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We can start by figuring out what the probability is for each die to add up to <math>7</math> if there is only <math>1</math> roll. We can quickly see that the probability is <math>\frac16</math> , as there are <math>6</math> ways to make <math>7</math> from the numbers on a die, and there are a total of <math>36</math> ways to add <math>2</math> numbers on a die. And since each time we roll the dice, we are adding to the probability, we can conclude that the total probability for <math>n</math> rolls would be <math>\frac16</math><math>n</math>. The smallest number that satisfies this is <math>\boxed {\textbf{(C) }4}</math>
+
On each roll, there is a <math>\frac 16</math> chance of rolling a sum of <math>7</math>. You would need <math>4</math> of these rolls to get <math>4 \cdot \frac 16,</math> which is larger than <math>\frac 12.</math> Therefore, the answer is <math>\boxed {\textbf{(C) }4}.</math>
  
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~ProProtractor
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~dbnl
  
 
==Video Solution (⚡️Just 4 min⚡️)==
 
==Video Solution (⚡️Just 4 min⚡️)==
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==Video Solution by Interstigation==
 
==Video Solution by Interstigation==
 
https://youtu.be/qT0hVzy7zeY
 
https://youtu.be/qT0hVzy7zeY
 +
 +
==Video Solution by TheBeautyofMath==
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https://youtu.be/Mi2AxPhnRno?t=207
  
 
== 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}}

Latest revision as of 23:27, 21 October 2024

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

Solution 3

We can start by figuring out what the probability is for each die to add up to $7$ if there is only $1$ roll. We can quickly see that the probability is $\frac16$, as there are $6$ ways to make $7$ from $2$ numbers on a die, and there are a total of $36$ ways to add $2$ numbers on a die. And since each time we roll the dice, we are adding to the probability, we can conclude that the total probability for rolling a sum of $7$ in $n$ rolls would be $\frac16$$n$. The smallest number that satisfies this is $\boxed {\textbf{(C) }4}$.

~mihikamishra

Solution 4 (Fakesolve)

On each roll, there is a $\frac 16$ chance of rolling a sum of $7$. You would need $4$ of these rolls to get $4 \cdot \frac 16,$ which is larger than $\frac 12.$ Therefore, the answer is $\boxed {\textbf{(C) }4}.$

~dbnl

Video Solution (⚡️Just 4 min⚡️)

https://youtu.be/rYyb3NCWBXk

~Education, the Study of Everything

Video Solution by Interstigation

https://youtu.be/qT0hVzy7zeY

Video Solution by TheBeautyofMath

https://youtu.be/Mi2AxPhnRno?t=207

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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