Difference between revisions of "2018 AMC 10B Problems/Problem 6"

Revision as of 18:57, 16 January 2023

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

Video Solution

https://youtu.be/ctQ3VbKAFBg

~savannahsolver

Video Solution by OmegaLearn

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

~ pi_is_3.14

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AMC 10 Problems and Solutions

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

@@ Line 7: / Line 7: @@
 == Solution 1 ==
-Notice that the only four ways such that <math>3</math> draws are required are <math>1,2</math>; <math>1,3</math>; <math>2,1</math>; and <math>3,1</math>. Notice that each of those cases has a <math>\frac{1}{5} \cdot \frac{1}{4}</math> chance, so the answer is <math>\frac{1}{5} \cdot \frac{1}{4} \cdot 4 = \frac{1}{5}</math>, or <math>\boxed{D}</math>.
+Notice that the only four ways such that <math>3</math> draws are required are <math>1,2</math>; <math>1,3</math>; <math>2,1</math>; and <math>1,3</math>. Notice that each of those cases has a <math>\frac{1}{5} \cdot \frac{1}{4}</math> chance, so the answer is <math>\frac{1}{5} \cdot \frac{1}{4} \cdot 4 = \frac{1}{5}</math>, or <math>\boxed{D}</math>.
+Jonathan Xu (pi_is_delicious_69420)
 == Solution 2 ==

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