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

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A box contains <math>5</math> chips, numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math>. Chips we drawn randomly one at a time without replacement until the sun of the values drawn exceeds <math>4</math>. What is the probability that <math>3</math> draws are required?
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== Problem ==
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A box contains <math>5</math> chips, numbered <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math>. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds <math>4</math>. What is the probability that <math>3</math> draws are required?
  
 
<math>\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}</math>
 
<math>\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}</math>
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== Solution 1 ==
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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>.
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Jonathan Xu (pi_is_delicious_69420)
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== Solution 1.1 ==
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Since our favorable outcomes are <math>1,2</math>; <math>1,3</math>; <math>2,1</math>; and <math>3,1</math>, we have 4 favorable outcomes. We have 20 total outcomes. Probability will be  <math>\frac{1}{5}</math>. Hence the answer is <math>\boxed{D}</math>.
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~east999
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== Solution 2 ==
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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 <math>1</math> to have 3 draws, otherwise <math>5</math> will be attainable in two or fewer draws. So the probability of getting a <math>1</math> is <math>\frac{1}{5}</math>. It is necessary to pull either a <math>2</math> or <math>3</math> on the next draw and the probability of that is <math>\frac{1}{2}</math>. But, the order of the draws can be switched so we get:
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<math>\frac{1}{5} \cdot \frac{1}{2} \cdot 2 = \frac{1}{5}</math>, or <math>\boxed {D}</math>
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~Soccer_JAMS
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== Solution 3 (Inequalities) ==
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We know that the first <math>2</math> draws must be <math>\le{3}</math> leaving us with with the only option being that <math>1</math> and <math>2</math> are chosen in either order. This means there is a probability of <math>\frac{1}{\binom52} \cdot 2 = \frac{1}{5}</math>, or <math>\boxed {D}</math>
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~MC_ADe
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==Video Solution (HOW TO THINK CRITICALLY!!!)==
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https://youtu.be/gtsDrM16J9U
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~Education, the Study of Everything
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==Video Solution==
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https://youtu.be/ctQ3VbKAFBg
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~savannahsolver
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== Video Solution ==
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https://youtu.be/wopflrvUN2c?t=20
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==See Also==
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{{AMC10 box|year=2018|ab=B|num-b=5|num-a=7}}
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{{MAA Notice}}

Latest revision as of 10:41, 5 November 2024

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 1.1

Since our favorable outcomes are $1,2$; $1,3$; $2,1$; and $3,1$, we have 4 favorable outcomes. We have 20 total outcomes. Probability will be $\frac{1}{5}$. Hence the answer is $\boxed{D}$.

~east999

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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