Question

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 $2?$

$\dfrac{2}{3}$

$\dfrac{19}{27}$

$\dfrac{59}{81}$

$\dfrac{61}{81}$

$\dfrac{7}{9}$

Solution:

Sort each die into low $\{1,2\},$ mid $\{3,4\},$ or high $\{5,6\};$ each has probability $\tfrac13,$ so the $3^4 = 81$ category patterns are equally likely.

We need at least one high die (a number greater than $4$ ) and at least two dice that are greater than $2$ (mid or high). Let $A$ be the event of at least one high and $B$ the event of at most one low die.

There are $2^4 = 16$ patterns with no high die, $9$ patterns with at most one non-low die, and $5$ patterns with neither a high die nor two non-low dice. By inclusion-exclusion the count of good patterns is $81 - 16 - 9 + 5 = 61.$

The probability is $\dfrac{61}{81}.$

Thus, the correct answer is D.

2022 AMC 12B Problem 12

Solution:

Problem 12 in Other Years