2021 AMC 10B Fall Problem 14

Question

Una rolls

6

standard

6

-sided dice simultaneously and calculates the product of the

6

numbers obtained. What is the probability that the product is divisible by

4?

\dfrac34

\dfrac{57}{64}

\dfrac{59}{64}

\dfrac{187}{192}

\dfrac{63}{64}

Count the complement: product not divisible by

4

The complement has either no factor of

2,

or exactly one factor of

2

Count the complement, where the product is not divisible by

4.

This happens if the product has no factor of

2,

or exactly one factor of

2.

All dice odd has probability

(\frac12)^6=\frac1{64}.

Exactly one factor of

2

means exactly one die is

2

or

6,

and the other five dice are odd. This has probability

6\cdot\frac26\cdot\left(\frac12\right)^5=\frac4{64}.

The complement has probability

\frac5{64},

so the desired probability is

1-\frac5{64}=\frac{59}{64}.

Thus, the answer is C .

Problem 14 in other years