2021 Fall AMC 12B Problems/Problem 11

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.

Problem

Una rolls $6$ standard $6$ -sided dice simultaneously and calculates the product of the $6{ }$ numbers obtained. What is the probability that the product is divisible by $4?$

$\textbf{(A)}\: \frac34\qquad\textbf{(B)} \: \frac{57}{64}\qquad\textbf{(C)} \: \frac{59}{64}\qquad\textbf{(D)} \: \frac{187}{192}\qquad\textbf{(E)} \: \frac{63}{64}$

Solution

We will use complementary counting to find the probability that the product is not divisible by $4$ . Then, we can find the probability that we want by subtracting this from 1. We split this into $2$ cases.

Case 1: The product is not divisible by $2$ .

We need every number to be odd, and since the chance we roll an odd number is $\frac12,$ our probability is $\left(\frac12\right)^6=\frac1{64}.$

Case 2: The product is divisible by $2$ , but not by $4$ .

We need $5$ numbers to be odd, and one to be divisible by $2$ , but not by $4$ . There is a $\frac12$ chance that an odd number is rolled, a $\frac13$ chance that we roll a number satisfying the second condition (only $2$ and $6$ work), and $6$ ways to choose the order in which the even number appears.

Our probability is $\left(\frac12\right)^5\left(\frac13\right)\cdot6=\frac1{16}.$

Therefore, the probability the product is not divisible by $4$ is $\frac1{64}+\frac1{16}=\frac{5}{64}$ .

Our answer is $1-\frac{5}{64}=\boxed{\textbf{(C)}\ \frac{59}{64}}$ .

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https://youtu.be/ucR90-j78hI

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https://youtu.be/KpHlKsUsEXo

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Video Solution by TheBeautyofMath

https://youtu.be/R7TwXgAGYuw?t=833

~IceMatrix

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.

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