# 2018 AMC 10B Problems/Problem 6

## Problem

A box contains $5$ chips, numbered $1$, $2$, $3$, $4$, and $5$. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds $4$. What is the probability that $3$ draws are required? $\textbf{(A)} \frac{1}{15} \qquad \textbf{(B)} \frac{1}{10} \qquad \textbf{(C)} \frac{1}{6} \qquad \textbf{(D)} \frac{1}{5} \qquad \textbf{(E)} \frac{1}{4}$

## Solution 1

Notice that the only four ways such that $3$ draws are required are $1,2$; $1,3$; $2,1$; and $1,3$. Notice that each of those cases has a $\frac{1}{5} \cdot \frac{1}{4}$ chance, so the answer is $\frac{1}{5} \cdot \frac{1}{4} \cdot 4 = \frac{1}{5}$, or $\boxed{D}$.

Jonathan Xu (pi_is_delicious_69420)

## Solution 2

We only have to analyze first two draws as that gives us insight on if third draw is necessary. Also, note that it is necessary to draw a $1$ in order to have 3 draws, otherwise $5$ will be attainable in two or less draws. So the probability of getting a $1$ is $\frac{1}{5}$. It is necessary to pull either a $2$ or $3$ on the next draw and the probability of that is $\frac{1}{2}$. But, the order of the draws can be switched so we get: $\frac{1}{5} \cdot \frac{1}{2} \cdot 2 = \frac{1}{5}$, or $\boxed {D}$

By: Soccer_JAMS

~savannahsolver

## Video Solution by OmegaLearn

~ pi_is_3.14

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 