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Difference between revisions of "2021 Fall AMC 12A Problems/Problem 7"

m (→‎Solution)
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==Solution==
 
==Solution==
−
The formula for expected values is <cmath>\text{Expected Value}=\sum(\text{Probability}\cdot\text{Outcome}).</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\cdot50 + \frac15\cdot20 + \frac15\cdot20 + \frac15\cdot5 + \frac15\cdot5 \\
+
t &= 50\cdot\frac15 + 20\cdot\frac15 + 20\cdot\frac15 + 5\cdot\frac15 + 5\cdot\frac15 \\
−
&= \frac15\cdot(50+20+20+5+5) \\
+
&= (50+20+20+5+5)\cdot\frac15 \\
−
&= \frac15\cdot100 \\
+
&= 100\cdot\frac15 \\
 
&= 20, \\
 
&= 20, \\
−
s &= \frac{50}{100}\cdot50 + \frac{20}{100}\cdot20 + \frac{20}{100}\cdot20 + \frac{5}{100}\cdot5 + \frac{5}{100}\cdot5 \\
+
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.

Revision as of 03:03, 29 March 2022

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.

Problem

A school has $100$ students and $5$ teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50, 20, 20, 5,$ and $5$. Let $t$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ 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 $t-s$?

$\textbf{(A)}\ {-}18.5 \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5$

Solution

The formula for expected values is \[\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).\] We have \begin{align*} t &= 50\cdot\frac15 + 20\cdot\frac15 + 20\cdot\frac15 + 5\cdot\frac15 + 5\cdot\frac15 \\ &= (50+20+20+5+5)\cdot\frac15 \\ &= 100\cdot\frac15 \\ &= 20, \\ 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 \\ &= 33.5. \end{align*} Therefore, the answer is $t-s=\boxed{\textbf{(B)}\ {-}13.5}.$

~MRENTHUSIASM

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. AMC logo.png

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