Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 SSMO Tiebreaker Round Problems/Problem 2"

(Created page with "==Problem== Bob is attempting to shoot a 3-point throw. Bob attempts the basket 97 times. Each time, Bob has a <math>35\%</math> chance of making the shot. If <math>S_1</math...")
 
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
 +
Firstly, the expected number of shots Bob makes is <math>\left(\frac{35}{100}\right)\cdot 97.</math> Now, the number of shots Bob is most likely to make is the largest integer <math>a</math> such that
 +
\begin{align*}
 +
\left(\frac{35}{100}\right)^a\left(\frac{65}{100}\right)^{97-a}\binom{97}{a}&\geq\left(\frac{35}{100}\right)^{a-1}\left(\frac{65}{100}\right)^{98-a}\binom{97}{a-1}\implies\\
 +
35(98-a)&\geq65(a)\implies\\
 +
3430&\geq100a\implies\\
 +
34&\geq a.
 +
\end{align*}
 +
So, Bob is most likely to make <math>34</math> shots. Therefore, the answer is <cmath>\left|3\left(\frac{35\cdot97}{100}-34\right)\right| = \frac{3}{20}\implies20+3 = \boxed{23}.</cmath>
 +
 +
~SMO_Team

Latest revision as of 14:48, 10 September 2025

Problem

Bob is attempting to shoot a 3-point throw. Bob attempts the basket 97 times. Each time, Bob has a $35\%$ chance of making the shot. If $S_1$ denotes the expected number of points Bob will make and $S_2$ the number of points Bob is most likely to make, then $|S_1-S_2| = \frac{m}{n},$ for relatively prime positive integers $m$ and $n.$ Find $m+n.$

Solution

Firstly, the expected number of shots Bob makes is $\left(\frac{35}{100}\right)\cdot 97.$ Now, the number of shots Bob is most likely to make is the largest integer $a$ such that \begin{align*} \left(\frac{35}{100}\right)^a\left(\frac{65}{100}\right)^{97-a}\binom{97}{a}&\geq\left(\frac{35}{100}\right)^{a-1}\left(\frac{65}{100}\right)^{98-a}\binom{97}{a-1}\implies\\ 35(98-a)&\geq65(a)\implies\\ 3430&\geq100a\implies\\ 34&\geq a. \end{align*} So, Bob is most likely to make $34$ shots. Therefore, the answer is \[\left|3\left(\frac{35\cdot97}{100}-34\right)\right| = \frac{3}{20}\implies20+3 = \boxed{23}.\]

~SMO_Team

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_SSMO_Tiebreaker_Round_Problems/Problem_2&oldid=260298"