2022 AMC 12B Problem 12

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}

Sort each die into low

\{1,2\},

mid

\{3,4\},

or high

\{5,6\},

each with probability

\tfrac13

You need at least one high die and at least two dice that are not low; count the good category patterns out of

3^4

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). The two bad events are having no high die and having at most one non-low die.

There are

2^4 = 16

patterns of the first kind,

9

of the second kind, and

5

in their intersection. 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.

Problem 12 in other years