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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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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>.
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It is obvious that the probability of getting at least two numbers greater than <math>2</math> is <math>1</math> if we have <math>2</math> numbers greater than <math>4</math>.
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Let us calculate the probability of getting at least two numbers greater than <math>2</math> if one die is greater than <math>4</math> by using complementary counting. We already have one die that is greater than <math>2</math>, and the probability that a die that is less than <math>5</math> is greater than <math>2</math> is <math>\frac{1}{2}</math>. Thus, our probability is <math>1 - \left(\frac{1}{2}\right)^3 = \frac{7}{8}</math>.
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Finally, our total probability is <math>\frac{11}{27} + \frac{7}{8}\cdot\frac{32}{81} = \boxed{\textbf{(D)}\ \frac{61}{81}}</math>
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~mathboy100
  
 
== See Also ==
 
== See Also ==

Revision as of 20:58, 19 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

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

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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