Difference between revisions of "2021 Fall AMC 12B Problems/Problem 11"

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==Video Solution by Interstigation==
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https://www.youtube.com/watch?v=G_31gUNwkzQ
  
 
==See Also==
 
==See Also==

Revision as of 13:29, 27 December 2021

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 11

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

~kingofpineapplz ~minor edits by stjwyl

Video Solution by Interstigation

https://www.youtube.com/watch?v=G_31gUNwkzQ

See Also

2021 Fall AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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