Difference between revisions of "2021 Fall AMC 12A Problems/Problem 7"
MRENTHUSIASM (talk | contribs) (The new edits are not in original wording.) |
|||
| (21 intermediate revisions by 7 users not shown) | |||
| Line 6: | Line 6: | ||
<math>\textbf{(A)}\ {-}18.5 \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5</math> | <math>\textbf{(A)}\ {-}18.5 \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5</math> | ||
| − | ==Solution | + | ==Solution== |
The formula for expected values is <cmath>\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).</cmath> | The formula for expected values is <cmath>\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).</cmath> | ||
We have | We have | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
| − | t &= \frac15\ | + | t &= 50\cdot\frac15 + 20\cdot\frac15 + 20\cdot\frac15 + 5\cdot\frac15 + 5\cdot\frac15 \\ |
| − | &= | + | &= (50+20+20+5+5)\cdot\frac15 \\ |
| − | &= \frac15 | + | &= 100\cdot\frac15 \\ |
&= 20, \\ | &= 20, \\ | ||
| − | s &= \frac{50}{100}\ | + | s &= 50\cdot\frac{50}{100} + 20\cdot\frac{20}{100} + 20\cdot\frac{20}{100} + 5\cdot\frac{5}{100} + 5\cdot\frac{5}{100} \\ |
&= 25 + 4 + 4 + 0.25 + 0.25 \\ | &= 25 + 4 + 4 + 0.25 + 0.25 \\ | ||
&= 33.5. | &= 33.5. | ||
| Line 22: | Line 22: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
| − | == Solution | + | ==Video Solution (Simple and Quick)== |
| − | + | https://youtu.be/nZBIoUS0mdk | |
| − | + | ~Education, the Study of Everything | |
| − | + | ||
| − | + | ||
| − | + | ==Video Solution by TheBeautyofMath== | |
| − | + | for AMC 10: https://youtu.be/ycRZHCOKTVk?t=789 | |
| − | |||
| − | |||
| − | |||
| − | |||
| − | + | for AMC 12: https://youtu.be/wlDlByKI7A8?t=157 | |
| + | ~IceMatrix | ||
| − | + | ==Video Solution by WhyMath== | |
| + | https://youtu.be/f7vhOCnvl0k | ||
| − | ~ | + | ~savannahsolver |
==See Also== | ==See Also== | ||
Latest revision as of 02:02, 13 July 2024
- The following problem is from both the 2021 Fall AMC 10A #10 and 2021 Fall AMC 12A #7, so both problems redirect to this page.
Contents
Problem
A school has 
students and 
teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are 
and . Let 
be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let 
be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is 
?
Solution
The formula for expected values is ![\[\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).\]](http://latex.artofproblemsolving.com/d/5/0/d508273b090899391634032a9b333fb87f84fe09.png)
We have
Therefore, the answer is 
~MRENTHUSIASM
Video Solution (Simple and Quick)
~Education, the Study of Everything
Video Solution by TheBeautyofMath
for AMC 10: https://youtu.be/ycRZHCOKTVk?t=789
for AMC 12: https://youtu.be/wlDlByKI7A8?t=157
~IceMatrix
Video Solution by WhyMath
~savannahsolver
See Also
| 2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 6 |
Followed by Problem 8 |
| 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 12 Problems and Solutions | |
| 2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
