Difference between revisions of "2022 AMC 10B Problems/Problem 12"
Mihikamishra (talk | contribs) (→Solution 3) |
m |
||
Line 29: | Line 29: | ||
==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
https://youtu.be/qT0hVzy7zeY | https://youtu.be/qT0hVzy7zeY | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/Mi2AxPhnRno?si=EdTtXNX-ynZAMyCL?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}} |
Revision as of 12:43, 21 June 2024
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
Video Solution by TheBeautyofMath
https://youtu.be/Mi2AxPhnRno?si=EdTtXNX-ynZAMyCL?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.