Difference between revisions of "2011 AMC 8 Problems/Problem 18"
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If dice <math>2</math> is <math>3</math>, then dice <math>1</math> can be <math>3, 4, 5,</math> or <math>6</math>. | If dice <math>2</math> is <math>3</math>, then dice <math>1</math> can be <math>3, 4, 5,</math> or <math>6</math>. | ||
− | In the first case, dice <math>1</math> had six values it could land on. In the second case, dice <math>1</math> had five values it could land on. In the third case, dice <math>1</math> had four values it could land on. If we keep going, we would eventually get the pattern <math>6+5+4+3+2+1</math> as the number of total values dice <math>1</math> could land on if it's greater or equal to dice <math>2</math>. Since there are <math>36</math> ways two dice could land, our answer is <math>\frac{21}{36}</math>, or <math>\boxed{\textbf{(D)} \frac{7}{12}}</math>. | + | In the first case, dice <math>1</math> had six values it could land on. In the second case, dice <math>1</math> had five values it could land on. In the third case, dice <math>1</math> had four values it could land on. If we keep going, we would eventually get the pattern <math>6+5+4+3+2+1=21</math> as the number of total values dice <math>1</math> could land on if it's greater or equal to dice <math>2</math>. Since there are <math>36</math> ways two dice could land, our answer is <math>\frac{21}{36}</math>, or <math>\boxed{\textbf{(D)} \frac{7}{12}}</math>. |
+ | |||
+ | == Video Solution by OmegaLearn == | ||
+ | https://youtu.be/6xNkyDgIhEE?t=139 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== Video Solution == | == Video Solution == | ||
https://youtu.be/lfrofNkOol0. Soo, DRMS, NM | https://youtu.be/lfrofNkOol0. Soo, DRMS, NM | ||
+ | ==Video Solution by WhyMath== | ||
+ | https://youtu.be/OWrMcByO3QQ | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=mYn6tNxrWBU | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=17|num-a=19}} | {{AMC8 box|year=2011|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 15:08, 17 December 2023
Contents
Problem
A fair sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?
Solution
There are ways to roll the two dice, and 6 of them result in two of the same number. Out of the remaining
ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are
ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is
Solution 2
If we list out some ways the first dice is greater than the second dice, we might see a pattern.
If dice is
, then dice
can be
or
.
If dice is
, then dice
can be
or
.
If dice is
, then dice
can be
or
.
In the first case, dice had six values it could land on. In the second case, dice
had five values it could land on. In the third case, dice
had four values it could land on. If we keep going, we would eventually get the pattern
as the number of total values dice
could land on if it's greater or equal to dice
. Since there are
ways two dice could land, our answer is
, or
.
Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=139
~ pi_is_3.14
Video Solution
https://youtu.be/lfrofNkOol0. Soo, DRMS, NM
Video Solution by WhyMath
~savannahsolver
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=mYn6tNxrWBU
See Also
2011 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.