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Difference between revisions of "1980 AHSME Problems/Problem 20"

(→‎Solution 1)
 
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\text{(E)}\ \text{none of these} </math>     
 
\text{(E)}\ \text{none of these} </math>     
 
    
 
    
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== Solution 1 ==
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Each of the <math>\binom{12}{6} = 924</math> possible combinations of <math>6</math> coins is equally likely to be selected.
  
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== Solution 1==
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If <math>6</math> dimes are selected, the total will be <math>60</math> cents.  There is only <math>\binom{6}{6} = 1</math> way to select <math>6</math> dimes.
 +
 
 +
If <math>5</math> dimes are selected, the total will be more than <math>50</math> cents.  There are <math>\binom{6}{5} = 6</math> ways to select the dimes, and <math>\binom{6}{1} = 6</math> ways to select the other coin, for a total of <math>6 \cdot 6 = 36</math> ways to select <math>5</math> dimes and <math>1</math> other coin.
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If <math>4</math> dimes are selected, the total will be at least <math>50</math> cents if and only if the other <math>2</math> coins are both nickels.  There are <math>\binom{6}{4} = 15</math> ways to select <math>4</math> dimes and <math>\binom{4}{2} = 6</math> ways to select <math>2</math> nickels, for a total of <math>15 \cdot 6 = 90</math> ways to select <math>4</math> dimes and <math>2</math> nickels.
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If <math>3</math> or fewer dimes are selected, the total will be less than <math>50</math> cents.
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Therefore, there are <math>1+36+90=127</math> ways to select <math>6</math> coins with a total of at least <math>50</math> cents, so the probability is <math>\boxed{(\textbf{C})\ \frac{127}{924}}</math>.
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-j314andrews
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== Solution 2==
 
We want the number of Successful Outcomes over the number of Total Outcomes. We want to calculate the total outcomes first. Since we have <math>12</math> coins and we need to choose <math>6</math>, we have <math>\binom{12}{6}</math> = <math>924</math> Total outcomes. For our successful outcomes, we can have <math>(1) 1</math> penny and <math>5</math> dimes, <math>2</math> nickels and <math>4</math> dimes, <math>1</math> nickel and <math>5</math> dimes, or <math>6</math> dimes.  
 
We want the number of Successful Outcomes over the number of Total Outcomes. We want to calculate the total outcomes first. Since we have <math>12</math> coins and we need to choose <math>6</math>, we have <math>\binom{12}{6}</math> = <math>924</math> Total outcomes. For our successful outcomes, we can have <math>(1) 1</math> penny and <math>5</math> dimes, <math>2</math> nickels and <math>4</math> dimes, <math>1</math> nickel and <math>5</math> dimes, or <math>6</math> dimes.  
  

Latest revision as of 17:29, 15 August 2025

Problem

A box contains $2$ pennies, $4$ nickels, and $6$ dimes. Six coins are drawn without replacement, with each coin having an equal probability of being chosen. What is the probability that the value of coins drawn is at least $50$ cents?

$\text{(A)} \ \frac{37}{924} \qquad \text{(B)} \ \frac{91}{924} \qquad \text{(C)} \ \frac{127}{924} \qquad \text{(D)}\ \frac{132}{924}\qquad \text{(E)}\ \text{none of these}$

Solution 1

Each of the $\binom{12}{6} = 924$ possible combinations of $6$ coins is equally likely to be selected.

If $6$ dimes are selected, the total will be $60$ cents. There is only $\binom{6}{6} = 1$ way to select $6$ dimes.

If $5$ dimes are selected, the total will be more than $50$ cents. There are $\binom{6}{5} = 6$ ways to select the dimes, and $\binom{6}{1} = 6$ ways to select the other coin, for a total of $6 \cdot 6 = 36$ ways to select $5$ dimes and $1$ other coin.

If $4$ dimes are selected, the total will be at least $50$ cents if and only if the other $2$ coins are both nickels. There are $\binom{6}{4} = 15$ ways to select $4$ dimes and $\binom{4}{2} = 6$ ways to select $2$ nickels, for a total of $15 \cdot 6 = 90$ ways to select $4$ dimes and $2$ nickels.

If $3$ or fewer dimes are selected, the total will be less than $50$ cents.

Therefore, there are $1+36+90=127$ ways to select $6$ coins with a total of at least $50$ cents, so the probability is $\boxed{(\textbf{C})\ \frac{127}{924}}$.

-j314andrews

Solution 2

We want the number of Successful Outcomes over the number of Total Outcomes. We want to calculate the total outcomes first. Since we have $12$ coins and we need to choose $6$, we have $\binom{12}{6}$ = $924$ Total outcomes. For our successful outcomes, we can have $(1) 1$ penny and $5$ dimes, $2$ nickels and $4$ dimes, $1$ nickel and $5$ dimes, or $6$ dimes.

For the case of $1$ penny and $5$ dimes, there are $\binom{6}{5}$ ways to choose the dimes and $2$ ways to choose the pennies. That is $6 \cdot 2 = 12$ successful outcomes. For the case of $2$ nickels and $4$ dimes, we have $\binom{6}{4}$ ways to choose the dimes and $\binom{4}{2}$ ways to choose the nickels. We have $15 \cdot 6$ = $90$ successful outcomes. For the case of $1$ nickel and $5$ dimes, we have $\binom{4}{1} \cdot \binom{6}{5} = 24$. Lastly, we have $6$ dimes and $0$ nickels, and $0$ pennies, so we only have one case. Therefore, we have $\dfrac {12 + 90 + 24 + 1}{924} = \dfrac{127}{924}$ = $\boxed{C}$

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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