Difference between revisions of "2022 AMC 10B Problems/Problem 12"
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<math>\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6</math> | <math>\textbf{(A) } 2 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 4 \qquad \textbf{(D) } 5 \qquad \textbf{(E) } 6</math> | ||
− | ==Solution== | + | ==Solution 1 (Complement)== |
+ | 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)^4<\frac12<\left(\frac56\right)^3,</math> the least integer <math>n</math> satisfying the inequality is <math>\boxed{\textbf{(C) } 4}.</math> | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | ==Solution 2 (99% Accurate Guesswork)== | ||
+ | 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 | ||
+ | |||
+ | |||
+ | ==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 | ||
+ | |||
+ | ==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 == | == 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 10:34, 19 August 2024
Contents
Problem
A pair of fair -sided dice is rolled
times. What is the least value of
such that the probability that the sum of the numbers face up on a roll equals
at least once is greater than
?
Solution 1 (Complement)
Rolling a pair of fair -sided dice, the probability of getting a sum of
is
Regardless what the first die shows, the second die has exactly one outcome to make the sum
We consider the complement: The probability of not getting a sum of
is
Rolling the pair of dice
times, the probability of getting a sum of
at least once is
Therefore, we have or
Since
the least integer
satisfying the inequality is
~MRENTHUSIASM
Solution 2 (99% Accurate Guesswork)
Let's try the answer choices. We can quickly find that when we roll dice, either the first and second sum to
, the first and third sum to
, or the second and third sum to
. There are
ways for the first and second dice to sum to
,
ways for the first and third to sum to
, and
ways for the second and third dice to sum to
. However, we overcounted (but not by much) so we can assume that the answer is
~Arcticturn
Solution 3
We can start by figuring out what the probability is for each die to add up to if there is only
roll. We can quickly see that the probability is
, as there are
ways to make
from
numbers on a die, and there are a total of
ways to add
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
in
rolls would be
. The smallest number that satisfies this is
~mihikamishra
Video Solution (⚡️Just 4 min⚡️)
~Education, the Study of Everything
Video Solution by Interstigation
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.