# Difference between revisions of "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 $3,1$. 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 $\boxed{D}$.

## Solution 2

Notice that only the first two draws are important, it doesn't matter what number we get third because no matter what combination of $3$ numbers is picked, the sum will always be greater than 5. 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

## Solution 3

We can use complementary probability. There is a $\frac{2}{5}$ chance of pulling either $4$ or $5$. In both cases, there is a 100% chance that we need not pull a third number. There is a $\frac{2}{5}$ chance of pulling either $3$ or $2$, for which there is a $\frac{3}{4}$ chance that we need not pull a third number, for this will only happen if $1$ is pulled next. Finally, if we pull a $1$ (for which the probability is $\frac{1}{5}$), there is a $\frac{1}{2}$ chance that we need not pull a third number, for this will happen if either $2$ or $3$ is pulled next.

Multiplying these fractions gives us the following expression: $\frac{2}{5} + \frac{2}{5} \cdot \frac{3}{4} + \frac{1}{5} \cdot \frac{1}{2}$

Therefore, the complementary probability is $\frac{4}{5},$ so the answer is $\boxed{\frac{1}{5}}$ or $\boxed{D}$.

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