Difference between revisions of "1980 AHSME Problems/Problem 20"
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\text{(E)}\ \text{none of these} </math> | \text{(E)}\ \text{none of these} </math> | ||
| + | == Solution 1 == | ||
| + | Each of the <math>\binom{12}{6} = 924</math> possible combinations of <math>6</math> coins is equally likely to be selected. | ||
| − | == Solution == | + | 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. |
| − | <math>\ | + | |
| + | 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. | ||
| + | |||
| + | 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. | ||
| + | |||
| + | If <math>3</math> or fewer dimes are selected, the total will be less than <math>50</math> cents. | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | -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 <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. | ||
| + | |||
| + | For the case of <math>1</math> penny and <math>5</math> dimes, there are <math>\binom{6}{5}</math> ways to choose the dimes and <math>2</math> ways to choose the pennies. That is <math>6 \cdot 2 = 12</math> successful outcomes. For the case of <math>2</math> nickels and <math>4</math> dimes, we have <math>\binom{6}{4}</math> ways to choose the dimes and <math>\binom{4}{2}</math> ways to choose the nickels. We have <math>15 \cdot 6</math> = <math>90</math> successful outcomes. For the case of <math>1</math> nickel and <math>5</math> dimes, we have <math>\binom{4}{1} \cdot \binom{6}{5} = 24</math>. Lastly, we have <math>6</math> dimes and <math>0</math> nickels, and <math>0</math> pennies, so we only have one case. Therefore, we have <math>\dfrac {12 + 90 + 24 + 1}{924} = \dfrac{127}{924}</math> = <math>\boxed{C}</math> | ||
== See also == | == See also == | ||
Latest revision as of 17:29, 15 August 2025
Contents
Problem
A box contains 
pennies, 
nickels, and 
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 
cents?
Solution 1
Each of the 
possible combinations of coins is equally likely to be selected.
If dimes are selected, the total will be 
cents. There is only 
way to select dimes.
If 
dimes are selected, the total will be more than cents. There are 
ways to select the dimes, and 
ways to select the other coin, for a total of 
ways to select dimes and 
other coin.
If dimes are selected, the total will be at least cents if and only if the other coins are both nickels. There are 
ways to select dimes and 
ways to select nickels, for a total of 
ways to select dimes and nickels.
If 
or fewer dimes are selected, the total will be less than cents.
Therefore, there are 
ways to select coins with a total of at least cents, so the probability is 
.
-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 
coins and we need to choose , we have 
= 
Total outcomes. For our successful outcomes, we can have 
penny and dimes, nickels and dimes, nickel and dimes, or dimes.
For the case of penny and dimes, there are 
ways to choose the dimes and ways to choose the pennies. That is 
successful outcomes. For the case of nickels and dimes, we have 
ways to choose the dimes and 
ways to choose the nickels. We have 
= 
successful outcomes. For the case of nickel and dimes, we have 
. Lastly, we have dimes and 
nickels, and pennies, so we only have one case. Therefore, we have 
= 
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
