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Difference between revisions of "2022 AMC 12B Problems/Problem 12"

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Therefore, the overall probability of meeting both conditions is <math>1 - \frac{16}{81} - \frac{4}{81} = \boxed{\textbf{(D)}\ \frac{61}{81}}</math>.
 
Therefore, the overall probability of meeting both conditions is <math>1 - \frac{16}{81} - \frac{4}{81} = \boxed{\textbf{(D)}\ \frac{61}{81}}</math>.
  
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== Solution 2 ==
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== Solution 2 (direct + complementary counting) ==
 
There are either <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math> dice that have values of more than <math>4</math>. The probability of getting <math>0</math> is <math>\left(\frac{2}{3}\right)^4 = \frac{16}{81}</math>, the probability of getting <math>1</math> is <math>4 \cdot \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^3 = \frac{32}{81}</math>, and the probability of getting <math>2</math> or greater is <math>1 - \frac{16}{81} - \frac{32}{81} = \frac{11}{27}</math>.  
 
There are either <math>0</math>, <math>1</math>, <math>2</math>, <math>3</math>, or <math>4</math> dice that have values of more than <math>4</math>. The probability of getting <math>0</math> is <math>\left(\frac{2}{3}\right)^4 = \frac{16}{81}</math>, the probability of getting <math>1</math> is <math>4 \cdot \left(\frac{1}{3}\right)\left(\frac{2}{3}\right)^3 = \frac{32}{81}</math>, and the probability of getting <math>2</math> or greater is <math>1 - \frac{16}{81} - \frac{32}{81} = \frac{11}{27}</math>.  
  

Revision as of 21:00, 20 November 2022

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 (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 + 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

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

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