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Difference between revisions of "2022 AMC 12B Problems/Problem 12"
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== Solution ==
 
== Solution ==
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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.
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For the first condition to be violated, all four dice must read <math>4</math> or less, which happens with probability <math>\left( \frac23 \right)^4 = \frac{16}{81}</math>.
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For the first condition to be met but the second condition to be violated, at least one of the dice must read greater than <math>4</math>, but less than two of the dice can read greater than <math>2</math>. Therefore, one of the four die must read <math>5</math> or <math>6</math>, while the remaining three dice must read <math>2</math> or less, which happens with probability <math>{4 \choose 1} \left(\frac13\right) \left(\frac13\right)^3 = 4 \cdot \frac13 \cdot \frac{1}{27} = \frac{4}{81}</math>.
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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>.
  
 
== See Also ==
 
== See Also ==

Revision as of 22:06, 17 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 $1 - \frac{16}{81} - \frac{4}{81} = \boxed{\textbf{(D)}\ \frac{61}{81}}$.

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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