Difference between revisions of "1980 AHSME Problems/Problem 20"
Hello12452 (talk | contribs) (→Solution) |
(→Solution 1) |
||
(6 intermediate revisions by 3 users not shown) | |||
Line 11: | Line 11: | ||
− | == Solution == | + | == Solution 1== |
− | <math>\ | + | 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 == | ||
{{AHSME box|year=1980|num-b=19|num-a=21}} | {{AHSME box|year=1980|num-b=19|num-a=21}} | ||
− | [[Category: Introductory | + | [[Category: Introductory Combinatorics Problems]] |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:25, 22 August 2022
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.