Difference between revisions of "2022 AMC 10B Problems/Problem 12"
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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> | 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)^ | + | 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 |
Revision as of 19:15, 17 November 2022
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
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
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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