From 0515e9167ea3f9a586ec9e10ed37772c5d90cf2d Mon Sep 17 00:00:00 2001
From: Trance-0 <60459821+Trance-0@users.noreply.github.com>
Date: Tue, 3 Dec 2024 13:54:15 -0600
Subject: [PATCH] Update CSE347_E2.md

---
 pages/CSE347/Exam_reviews/CSE347_E2.md | 15 ++++++++++-----
 1 file changed, 10 insertions(+), 5 deletions(-)

diff --git a/pages/CSE347/Exam_reviews/CSE347_E2.md b/pages/CSE347/Exam_reviews/CSE347_E2.md
index e444a62..95be0c2 100644
--- a/pages/CSE347/Exam_reviews/CSE347_E2.md
+++ b/pages/CSE347/Exam_reviews/CSE347_E2.md
@@ -114,11 +114,16 @@ $$
 - Assume we run the algorithm $k$ times, and the probability of success is $\frac{1}{2} + \epsilon$.
 - The probability that all trials fail is at most $(1-\epsilon)^k$.
 - The majority vote of $k$ runs is wrong is the same as probability that more than $\frac{k}{2}+1$ trials fail.
-- So, the probability is  
-  $$\begin{aligned}
-  Pr[\text{majority fails}] &=\sum_{i=\frac{k}{2}+1}^{k}\binom{k}{i}(\frac{1}{2}-\epsilon)^i(\frac{1}{2}+\epsilon)^{k-i}\\
-  &= \binom{k}{\frac{k}{2}+1}(\frac{1}{2}-\epsilon)^{\frac{k}{2}+1}
-  \end{aligned}$$
+- So, the probability is 
+
+$$
+\begin{aligned}
+Pr[\text{majority fails}] &=\sum_{i=\frac{k}{2}+1}^{k}\binom{k}{i}(\frac{1}{2}-\epsilon)^i(\frac{1}{2}+\epsilon)^{k-i}\\
+&= \binom{k}{\frac{k}{2}+1}(\frac{1}{2}-\epsilon)^{\frac{k}{2}+1}
+\end{aligned}
+$$
+
+
 - If we want this probability to be at most $p$, we can just solve for $k$ in the inequality make it less than some $\delta$. Then we solve for $k$ in the inequality $\binom{k}{\frac{k}{2}+1}(\frac{1}{2}-\epsilon)^{\frac{k}{2}+1} \leq \delta$.
 
 ## Online Algorithms