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Difference between revisions of "2005 AMC 10B Problems/Problem 21"

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Hence, <math>\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=162\implies \boxed{A}</math>.
 
Hence, <math>\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=162\implies \boxed{A}</math>.
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\(\text{Let <math>{\color{magenta}{m}} = {\color{magenta}{magenta}}</math> } \\ \text{Let <math>{\color{cyan}{c}} = {\color{cyan}{cyan}}</math> } \\ \text{Let <math>{\color{black}{b}} = {\color{black}{black}}</math> } \\ \text{Let <math>{\color{grey}{w}} = {\color{grey}{white}}</math> } \)
 
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The Set is\(\{ {\color{magenta}{m_1}},{\color{magenta}{m_2}}, {\color{cyan}{c_1}},{\color{cyan}{c_2}}, {\color{black}{b_1}},{\color{black}{b_2}}, {\color{grey}{w_1}},{\color{grey}{w_2}} \} \)
 
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The number of all the possibilities is \(^8C_4=\dbinom{8}{4} = \mathbf{70 }\)
 
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\(\begin{array}{|rcll|} \hline && \dfrac{ \dbinom{ {\color{magenta}{2} } }{ 2 }\times \left[ \dbinom{{\color{cyan}{2}}}{1}\dbinom{{\color{black}{2}}}{1}\dbinom{\color{grey}{2}}{0} +\dbinom{{\color{cyan}{2}}}{1}\dbinom{{\color{black}{2}}}{0}\dbinom{\color{grey}{2}}{1} +\dbinom{{\color{cyan}{2}}}{0}\dbinom{{\color{black}{2}}}{1}\dbinom{\color{grey}{2}}{1} \right] \\ + \dbinom{ {\color{cyan}{2} } }{ 2 }\times \left[ \dbinom{{\color{magenta}{2}}}{1}\dbinom{{\color{black}{2}}}{1}\dbinom{\color{grey}{2}}{0} +\dbinom{{\color{magenta}{2}}}{1}\dbinom{{\color{black}{2}}}{0}\dbinom{\color{grey}{2}}{1} +\dbinom{{\color{magenta}{2}}}{0}\dbinom{{\color{black}{2}}}{1}\dbinom{\color{grey}{2}}{1} \right] \\ + \dbinom{ {\color{black}{2} } }{ 2 }\times \left[ \dbinom{{\color{cyan}{2}}}{1}\dbinom{{\color{magenta}{2}}}{1}\dbinom{\color{grey}{2}}{0} +\dbinom{{\color{cyan}{2}}}{1}\dbinom{{\color{magenta}{2}}}{0}\dbinom{\color{grey}{2}}{1} +\dbinom{{\color{cyan}{2}}}{0}\dbinom{{\color{magenta}{2}}}{1}\dbinom{\color{grey}{2}}{1} \right] \\ + \dbinom{ {\color{grey}{2} } }{ 2 }\times \left[ \dbinom{{\color{cyan}{2}}}{1}\dbinom{{\color{black}{2}}}{1}\dbinom{\color{magenta}{2}}{0} +\dbinom{{\color{cyan}{2}}}{1}\dbinom{{\color{black}{2}}}{0}\dbinom{\color{magenta}{2}}{1} +\dbinom{{\color{cyan}{2}}}{0}\dbinom{{\color{black}{2}}}{1}\dbinom{\color{magenta}{2}}{1} \right] } {70} \\\\ &=& \dfrac{ 1\times \left[ 2\cdot 2\cdot 1 + 2\cdot 1\cdot 2 + 1\cdot 2\cdot 2 \right] \\ + 1\times \left[ 2\cdot 2\cdot 1 + 2\cdot 1\cdot 2 + 1\cdot 2\cdot 2 \right] \\ + 1\times \left[ 2\cdot 2\cdot 1 + 2\cdot 1\cdot 2 + 1\cdot 2\cdot 2 \right] \\ + 1\times \left[ 2\cdot 2\cdot 1 + 2\cdot 1\cdot 2 + 1\cdot 2\cdot 2 \right] } {70} \\\\ &=& \dfrac{ ( 4+4+4 ) + (4+4+4) + (4+4+4) + (4+4+4) } {70} \\ &=& \dfrac{ 12 + 12 + 12 + 12 } {70} \\ &=& \dfrac{ 4\cdot 12 } {70} \\\\ &=& \dfrac{ 2\cdot 12 } {35} \\\\ &\mathbf{=}&\mathbf{ \dfrac{ 24 } {35} } \\ \hline \end{array}\)
 
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The probability that she has exactly one pair of socks with the same color is \(\mathbf{ \dfrac{ 24 } {35} }\)
 
  
 
==See Also==
 
==See Also==

Revision as of 20:19, 18 August 2020

Problem

Forty slips are placed into a hat, each bearing a number $1$, $2$, $3$, $4$, $5$, $6$, $7$, $8$, $9$, or $10$, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b \neq a$. What is the value of $q/p$?

$\mathrm{(A)} 162 \qquad \mathrm{(B)} 180 \qquad \mathrm{(C)} 324 \qquad \mathrm{(D)} 360 \qquad \mathrm{(E)} 720$

Solution 1 (where the order of drawing slips matters)

There are $10$ ways to determine which number to pick. There are $4!$ ways to then draw those four slips with that number, and $40 \cdot 39 \cdot 38 \cdot 37$ total ways to draw four slips. Thus $p = \frac{10\cdot 4!}{40 \cdot 39 \cdot 38 \cdot 37}$.

There are ${10 \choose 2} = 45$ ways to determine which two numbers to pick for the second probability. There are ${4 \choose 2} = 6$ ways to arrange the order which we draw the non-equal slips, and in each order there are $4 \times 3 \times 4 \times 3$ ways to pick the slips, so $q = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{40 \cdot 39 \cdot 38 \cdot 37}$.

Hence, the answer is $\frac{q}{p} = \frac{45 \cdot 6 \cdot 4^2 \cdot 3^2}{10\cdot 4!} = \boxed{\ \mathbf{(A)}162}$.

Solution 2 (order does not matter)

For probability $p$, there are $\binom{10}{1}=10$ ways to choose the card you want to show up $4$ times.

Hence, the probability is $\frac{10}{\binom{40}{4}}$.

For probability $q$, there are $\binom{10}{2}=45$ ways to choose the $2$ numbers you want to show up twice. There are $\binom{4}{2}\cdot\binom{4}{2}$ to pick which cards you want out of the $4$ of each.

Hence, the probability is $\frac{45\cdot6\cdot6}{\binom{40}{4}}$

Hence, $\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=162\implies \boxed{A}$.

See Also

2005 AMC 10B (Problems • Answer Key • Resources)
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Problem 20 		Followed by
Problem 22
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All AMC 10 Problems and Solutions

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