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Difference between revisions of "2018 AMC 10B Problems/Problem 6"
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Notice that the only two ways such that more than <math>2</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 two ways such that more than <math>2</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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== Solution 2 ==
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Notice that only the first two draws are important, it doesn't matter what number we get third because no matter what combination of <math>3</math> numbers is  picked, the sum will always be greater than 5.Also, note that it is necessary to draw a <math>1</math> in order to have 3 draws, otherwise <math>5</math> will be attainable in two or less 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>
  
 
==See Also==
 
==See Also==

Revision as of 16:06, 16 February 2018

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

Notice that the only two ways such that more than $2$ 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 $\boxed {D}$

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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