Difference between revisions of "2021 Fall AMC 12B Problems/Problem 11"
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{{duplicate|[[2021 Fall AMC 10B Problems#Problem 14|2021 Fall AMC 10B #14]] and [[2021 Fall AMC 12B Problems#Problem 11|2021 Fall AMC 12B #11]]}} | {{duplicate|[[2021 Fall AMC 10B Problems#Problem 14|2021 Fall AMC 10B #14]] and [[2021 Fall AMC 12B Problems#Problem 11|2021 Fall AMC 12B #11]]}} | ||
| − | ==Problem | + | ==Problem== |
Una rolls <math>6</math> standard <math>6</math>-sided dice simultaneously and calculates the product of the <math>6{ }</math> numbers obtained. What is the probability that the product is divisible by <math>4?</math> | Una rolls <math>6</math> standard <math>6</math>-sided dice simultaneously and calculates the product of the <math>6{ }</math> numbers obtained. What is the probability that the product is divisible by <math>4?</math> | ||
| Line 6: | Line 6: | ||
==Solution== | ==Solution== | ||
| − | We will | + | We will use complementary counting to find the probability that the product is not divisible by <math>4</math>. Then, we can find the probability that we want by subtracting this from 1. We split this into <math>2</math> cases. |
Case 1: The product is not divisible by <math>2</math>. | Case 1: The product is not divisible by <math>2</math>. | ||
We need every number to be odd, and since the chance we roll an odd number is <math>\frac12,</math> our probability is <math>\left(\frac12\right)^6=\frac1{64}.</math> | We need every number to be odd, and since the chance we roll an odd number is <math>\frac12,</math> our probability is <math>\left(\frac12\right)^6=\frac1{64}.</math> | ||
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| − | |||
Case 2: The product is divisible by <math>2</math>, but not by <math>4</math>. | Case 2: The product is divisible by <math>2</math>, but not by <math>4</math>. | ||
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Our answer is <math>1-\frac{5}{64}=\boxed{\textbf{(C)}\ \frac{59}{64}}</math>. | Our answer is <math>1-\frac{5}{64}=\boxed{\textbf{(C)}\ \frac{59}{64}}</math>. | ||
| − | ~kingofpineapplz | + | ~kingofpineapplz (Most of Solution) |
| + | |||
| + | ~stjwyl (Edits) | ||
| + | |||
| + | ==Video Solution by Interstigation== | ||
| + | https://www.youtube.com/watch?v=G_31gUNwkzQ | ||
| + | |||
| + | ==Video Solution (Just 4 min!)== | ||
| + | https://youtu.be/ucR90-j78hI | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/KpHlKsUsEXo | ||
| + | |||
| + | ~savannahsolver | ||
| + | ==Video Solution by TheBeautyofMath== | ||
| + | https://youtu.be/R7TwXgAGYuw?t=833 | ||
| + | |||
| + | ~IceMatrix | ||
==See Also== | ==See Also== | ||
Latest revision as of 18:47, 1 November 2023
- The following problem is from both the 2021 Fall AMC 10B #14 and 2021 Fall AMC 12B #11, so both problems redirect to this page.
Contents
Problem
Una rolls 
standard -sided dice simultaneously and calculates the product of the 
numbers obtained. What is the probability that the product is divisible by 
Solution
We will use complementary counting to find the probability that the product is not divisible by 
. Then, we can find the probability that we want by subtracting this from 1. We split this into 
cases.
Case 1: The product is not divisible by .
We need every number to be odd, and since the chance we roll an odd number is 
our probability is 
Case 2: The product is divisible by , but not by .
We need 
numbers to be odd, and one to be divisible by , but not by . There is a 
chance that an odd number is rolled, a 
chance that we roll a number satisfying the second condition (only and work), and ways to choose the order in which the even number appears.
Our probability is 
Therefore, the probability the product is not divisible by is 
.
Our answer is 
.
~kingofpineapplz (Most of Solution)
~stjwyl (Edits)
Video Solution by Interstigation
https://www.youtube.com/watch?v=G_31gUNwkzQ
Video Solution (Just 4 min!)
~Education, the Study of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by TheBeautyofMath
https://youtu.be/R7TwXgAGYuw?t=833
~IceMatrix
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 | ||
| 2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 10 |
Followed by Problem 12 |
| 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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
