Difference between revisions of "2021 Fall AMC 12A Problems/Problem 7"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | |||
+ | ==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 03:02, 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
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.