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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Let <math>{\color{magenta}{m}} = {\color{magenta}{magenta}}</math> Let <math>{\color{cyan}{c}} = {\color{cyan}{cyan}}</math> Let <math>{\color{black}{b}} = {\color{black}{black}}</math> Let <math>{\color{grey}{w}} = {\color{grey}{white}}</math> 
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The Set is{m1,m2,c1,c2,b1,b2,w1,w2}
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The number of all the possibilities is 8C4=(84)=70
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(22)×[(21)(21)(20)+(21)(20)(21)+(20)(21)(21)]+(22)×[(21)(21)(20)+(21)(20)(21)+(20)(21)(21)]+(22)×[(21)(21)(20)+(21)(20)(21)+(20)(21)(21)]+(22)×[(21)(21)(20)+(21)(20)(21)+(20)(21)(21)]70=1×[221+212+122]+1×[221+212+122]+1×[221+212+122]+1×[221+212+122]70=(4+4+4)+(4+4+4)+(4+4+4)+(4+4+4)70=12+12+12+1270=41270=21235=2435
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The probability that she has exactly one pair of socks with the same color is 2435
  
 
==See Also==
 
==See Also==

Revision as of 20:18, 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, $\frac{q}{p}=\frac{45\cdot6\cdot6}{10}=9\cdot6\cdot3=162\implies \boxed{A}$.

\(\text{Let ${\color{magenta}{m}} = {\color{magenta}{magenta}}$ } \ \text{Let ${\color{cyan}{c}} = {\color{cyan}{cyan}}$ } \ \text{Let ${\color{black}{b}} = {\color{black}{black}}$ } \ \text{Let ${\color{grey}{w}} = {\color{grey}{white}}$ (Error compiling LaTeX. Unknown error_msg) } \)


The Set is{m1,m2,c1,c2,b1,b2,w1,w2}


The number of all the possibilities is 8C4=(84)=70


(22)×[(21)(21)(20)+(21)(20)(21)+(20)(21)(21)]+(22)×[(21)(21)(20)+(21)(20)(21)+(20)(21)(21)]+(22)×[(21)(21)(20)+(21)(20)(21)+(20)(21)(21)]+(22)×[(21)(21)(20)+(21)(20)(21)+(20)(21)(21)]70=1×[221+212+122]+1×[221+212+122]+1×[221+212+122]+1×[221+212+122]70=(4+4+4)+(4+4+4)+(4+4+4)+(4+4+4)70=12+12+12+1270=41270=21235=2435


The probability that she has exactly one pair of socks with the same color is 2435

See Also

2005 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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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