Difference between revisions of "2022 AMC 10B Problems/Problem 12"
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==Solution 3== | ==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 | + | 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 <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> |
~ProProtractor | ~ProProtractor |
Revision as of 13:50, 12 November 2023
Contents
[hide]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 rolls would be . The smallest number that satisfies this is
~ProProtractor
Video Solution (⚡️Just 4 min⚡️)
~Education, the Study of Everything
Video Solution by Interstigation
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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