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

(I added a solution that uses complementary probability.)
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== Solution 2 ==
 
== 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:
+
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 <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:
  
 
<math>\frac{1}{5} \cdot \frac{1}{2} \cdot 2 = \frac{1}{5}</math>, or <math>\boxed {D}</math>
 
<math>\frac{1}{5} \cdot \frac{1}{2} \cdot 2 = \frac{1}{5}</math>, or <math>\boxed {D}</math>
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== Solution 3 ==
 
== Solution 3 ==
  
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We can use complementary probability. There is a <math>\frac{2}{5}</math> chance of pulling either <math>4</math> or <math>5</math>. In both cases, there is a <math>100%</math> chance that we need not pull a third number. There is a <math>\frac{2}{5}</math> chance of pulling either <math>3</math> or <math>2</math>, for which there is a <math>\frac{3}{4}</math> chance that we need not pull a third number, for this will only happen if <math>1</math> is pulled next. Finally, if we pull a <math>1</math> (for which the probability is <math>\frac{1}{5}</math>), there is a <math>\frac{1}{2}</math> chance that we need not pull a third number, for this will happen if either <math>2</math> or <math>3</math> is pulled next.
+
We can use complementary probability. There is a <math>\frac{2}{5}</math> chance of pulling either <math>4</math> or <math>5</math>. In both cases, there is a 100% chance that we need not pull a third number. There is a <math>\frac{2}{5}</math> chance of pulling either <math>3</math> or <math>2</math>, for which there is a <math>\frac{3}{4}</math> chance that we need not pull a third number, for this will only happen if <math>1</math> is pulled next. Finally, if we pull a <math>1</math> (for which the probability is <math>\frac{1}{5}</math>), there is a <math>\frac{1}{2}</math> chance that we need not pull a third number, for this will happen if either <math>2</math> or <math>3</math> is pulled next.
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Multiplying these fractions gives us the following expression:
 
Multiplying these fractions gives us the following expression:
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<math>\frac{2}{5} + \frac{2}{5} \cdot \frac{3}{4} + \frac{1}{5} \cdot \frac{1}{2}</math>
 
<math>\frac{2}{5} + \frac{2}{5} \cdot \frac{3}{4} + \frac{1}{5} \cdot \frac{1}{2}</math>
  
−
Therefore, the complementary probability is <math>\frac{4}{5}</math>, so the answer is <math>\frac{1}{5}</math>.
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Therefore, the complementary probability is <math>\frac{4}{5},</math> so the answer is <math>\boxed{\frac{1}{5}}</math> or <math>\boxed{D}</math>.
 +
 
 +
==Video Solution==
 +
https://youtu.be/ctQ3VbKAFBg
 +
 
 +
~savannahsolver
 +
 
 +
== Video Solution ==
 +
https://youtu.be/wopflrvUN2c?t=20
 +
 
 +
~ pi_is_3.14
  
 
==See Also==
 
==See Also==

Revision as of 16:33, 1 September 2021

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

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

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 $\boxed{D}$.

Video Solution

https://youtu.be/ctQ3VbKAFBg

~savannahsolver

Video Solution

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

~ pi_is_3.14

See Also

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

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