From f364328e02d5655460f5c4a941be2bd7ccaae919 Mon Sep 17 00:00:00 2001
From: stevee-bee <77158150+stevee-bee@users.noreply.github.com>
Date: Wed, 17 Mar 2021 04:31:07 +0800
Subject: [PATCH] clarifying the transition between statements

I was unable to understand the jump between the above statements. I posted the question in the related Coursera forum and Alan Berger (mentor) was kind enough to break it down for me. Using Alan's explanation, I offer this suggestion to further enhance the swirl lesson between the statements at 26% and 29%.
---
 Statistical_Inference/ConditionalProbability/lesson | 6 ++++++
 1 file changed, 6 insertions(+)

diff --git a/Statistical_Inference/ConditionalProbability/lesson b/Statistical_Inference/ConditionalProbability/lesson
index bbcda7c2..d7c6826c 100644
--- a/Statistical_Inference/ConditionalProbability/lesson
+++ b/Statistical_Inference/ConditionalProbability/lesson
@@ -58,6 +58,12 @@
 - Class: text
   Output: Suppose we don't know P(A) itself, but only know its conditional probabilities, that is, the probability that it occurs if B occurs and the probability that it occurs if B doesn't occur. These are  P(A|B) and P(A|~B), respectively. We use ~B to represent 'not B' or 'B complement'. 
 
+- Class: text
+  Output: Since B and ~B are by definition disjoint sets, we can add the probabilities of the two subsets, P(A|B) and P(A|~B), to give the probability of A. 
+
+- Class: text
+  Output: Then we can rewrite P(A) = P(A|B) + P(A|~B) using our initial conditional probability statement, P(A|B) = P(A&B) / P(B), as well as the same for P(A|~B). 
+
 - Class: text
   Output: We can then express P(A) = P(A|B) * P(B) + P(A|~B) * P(~B) and substitute this is into the denominator of Bayes' Formula.