Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2022 AMC 12B Problems/Problem 12

Revision as of 18:38, 9 January 2023 by MRENTHUSIASM (talk | contribs) (→‎Solution 3 (complementary counting + PIE))

Problem

Kayla rolls four fair $6$-sided dice. What is the probability that at least one of the numbers Kayla rolls is greater than $4$ and at least two of the numbers she rolls are greater than $2$?

$\textbf{(A)}\ \frac{2}{3} \qquad \textbf{(B)}\ \frac{19}{27} \qquad \textbf{(C)}\ \frac{59}{81} \qquad \textbf{(D)}\ \frac{61}{81} \qquad \textbf{(E)}\ \frac{7}{9}$

Solution 1 (Complementary Counting)

We will subtract from one the probability that the first condition is violated and the probability that only the second condition is violated, being careful not to double-count the probability that both conditions are violated.

For the first condition to be violated, all four dice must read $4$ or less, which happens with probability $\left( \frac23 \right)^4 = \frac{16}{81}$.

For the first condition to be met but the second condition to be violated, at least one of the dice must read greater than $4$, but less than two of the dice can read greater than $2$. Therefore, one of the four die must read $5$ or $6$, while the remaining three dice must read $2$ or less, which happens with probability ${4 \choose 1} \left(\frac13\right) \left(\frac13\right)^3 = 4 \cdot \frac13 \cdot \frac{1}{27} = \frac{4}{81}$.

Therefore, the overall probability of meeting both conditions is $1 - \frac{16}{81} - \frac{4}{81} = \boxed{\textbf{(D)}\ \frac{61}{81}}$.

Solution 2 (Direct and Complementary Counting)

There are either $0$, $1$, $2$, $3$, or $4$ dice that have values of more than $4$. The probability of getting $0$ is $\left(\frac{2}{3}\right)^4 = \frac{16}{81}$, the probability of getting $1$ is $4 \cdot \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^3 = \frac{32}{81}$, and the probability of getting $2$ or greater is $1 - \frac{16}{81} - \frac{32}{81} = \frac{11}{27}$.

It is obvious that the probability of getting at least two numbers greater than $2$ is $1$ if we have $2$ numbers greater than $4$.

Let us calculate the probability of getting at least two numbers greater than $2$ if one die is greater than $4$ by using complementary counting. We already have one die that is greater than $2$, and the probability that a die that is less than $5$ is greater than $2$ is $\frac{1}{2}$. Thus, our probability is $1 - \left(\frac{1}{2}\right)^3 = \frac{7}{8}$.

Finally, our total probability is $\frac{11}{27} + \frac{7}{8}\cdot\frac{32}{81} = \boxed{\textbf{(D)}\ \frac{61}{81}}$

~mathboy100

Solution 3 (Complementary Counting and PIE)

We use Solution 1's idea of complementary counting. However, we will find the probability that the first condition is violated and the probability that the second condition is violated, then subtract the overlap (first and second condition both violated). We can due this because of PIE.

As in Solution 1, the probability that the $\textbf{first condition}$ is violated is $\left(\frac{2}{3}\right)^4 = \frac{16}{81}.$

The probability that the $\textbf{second condition}$ is violated (regardless of the first condition) can be broken into two cases.

$\textbf{Case 1: exactly one number rolled is greater than 2}$

Then, there are four choices for which die is the one greater than 2, $\frac{2}{3}$ chance for whether it reads 3, 4, 5, or 6, and $\left(\frac{1}{3}\right)^3 = \frac{1}{27}$ for the other three die that must read 1 or 2. The probability is then $4\left(\frac{2}{3}\right)\left(\frac{1}{27}\right) = \frac{8}{81}.$

$\textbf{Case 2: exactly zero numbers rolled are greater than 2}$

Then, all four dice are either 1 or 2, so the probability of this is $\left(\frac{1}{3}\right)^4 = \frac{1}{81}.$

Now, we look at the $\textbf{overlap}$. In other words, we want to find the probability none of the rolls are greater than $4$ and less than two of the rolls are greater than $2$. We split into the same two cases again.

$\textbf{Case 1: exactly one number rolled is greater than 2 (and none of the rolls are greater than 4)}$

There are still 4 choices for this die, and it must be 3 or 4, with probability $\frac{1}{3}$. The other must read 1 or 2, with probability $\frac{1}{27},$ so this case yields $4\left(\frac{1}{3}\right)\left(\frac{1}{27}\right) = \frac{4}{81}.$

$\textbf{Case 2: exactly zero numbers rolled are greater than 2 (and none of the rolls are greater than 4)}$

The first statement implies the second statement, so this case yields $\frac{1}{81}$ as from our previous Case 2.

Now we've got everything! Our answer is $1 - (\frac{16}{81} + (\frac{8}{81} + \frac{1}{81}) - (\frac{4}{81} + \frac{1}{81})) = \boxed{\textbf{(D)}\ \frac{61}{81}}.$

~sirswagger21

See Also

2022 AMC 12B (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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12B_Problems/Problem_12&oldid=186166"