Question

A bag contains four pieces of paper, each labeled with one of the digits $1,$ $2,$ $3$ or $4,$ with no repeats. Three of these pieces are drawn, one at a time without replacement, to construct a three-digit number. What is the probability that the three-digit number is a multiple of $3?$

$\dfrac{1}{4}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{2}{3}$

$\dfrac{3}{4}$

Solution:

Recall that a number is divisible by $3$ if the sum of its digits is divisible by $3$ .

The only triples of distinct digits from $1,2,3,4$ whose sum is divisible by $3$ are $(1,2,3)$ and $(2,3,4)$ .

This means the constructed number is a multiple of $3$ exactly when either $1$ or $4$ is left in the bag.

Each of these events has probability $\dfrac{1}{4}$ , for a total probability of $2\cdot\dfrac{1}{4}=\dfrac{1}{2}$ .

Thus, C is the correct answer.

2007 AMC 8 Problem 24

Solution:

Problem 24 in Other Years