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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 $\boxed{D}$.

Jonathan Xu (pi_is_delicious_69420)

Solution 2

We only have to analyze the first two draws as that gives us insight into if a third draw is necessary. Also, note that it is necessary to draw a $1$ to have 3 draws, otherwise $5$ will be attainable in two or fewer 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 $\boxed {D}$

~Soccer_JAMS

Solution 3 (Inequalities)

We know that the first $2$ draws must be $\le{3}$ leaving us with with the only option being that $1$ and $2$ are chosen in either order. This means there is a probability of $\frac{1}{\binom52} \cdot 2 = \frac{1}{5}$, or $\boxed {D}$

~MC_ADe

Video Solution (HOW TO THINK CRITICALLY!!!)

https://youtu.be/gtsDrM16J9U

~Education, the Study of Everything

Video Solution

https://youtu.be/ctQ3VbKAFBg

~savannahsolver

Video Solution

https://youtu.be/wopflrvUN2c?t=20

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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