2022 AMC 10B Problems/Problem 12
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 |
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