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

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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>.
 
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>.
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==Solution 4==
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We see that multiplying <cmath>5, 4, and 3</cmath>, there are a total of <cmath>60</cmath> ways to draw the chips. For this, the first <cmath>2</cmath> draws must not exceed <cmath>4</cmath>. Now we have ordered pairs for the possible first and second draws which are: <cmath>(1, 2), (2, 1), (1, 3), and (3, 1).</cmath>For each of the ordered pairs we have the probability being <math>\frac{3}{60}</math>. This results in the probability <math>\frac{12}{60} = frac{1}{5} or </math>\boxed{D}$.
  
 
==See Also==
 
==See Also==

Revision as of 09:39, 9 April 2020

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

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 $\boxed {D}$

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 $\boxed{D}$.


Solution 4

We see that multiplying \[5, 4, and 3\], there are a total of \[60\] ways to draw the chips. For this, the first \[2\] draws must not exceed \[4\]. Now we have ordered pairs for the possible first and second draws which are: \[(1, 2), (2, 1), (1, 3), and (3, 1).\]For each of the ordered pairs we have the probability being $\frac{3}{60}$. This results in the probability $\frac{12}{60} = frac{1}{5} or$\boxed{D}$.

See Also

2018 AMC 10B (ProblemsAnswer KeyResources)
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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