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

Revision as of 22:00, 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 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 + 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

@@ Line 9: / Line 9: @@
   \textbf{(E)}\ \frac{7}{9}</math>
-== Solution (complementary counting) ==
+== 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.

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

Page

Toolbox

Search

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

Revision as of 22:00, 20 November 2022

Contents

Problem

Solution 1 (complementary counting)

Solution 2 (direct + complementary counting)

See Also