From 61555fc51ca079e2d9884966f83b81b3b976be55 Mon Sep 17 00:00:00 2001 From: eggy Date: Mon, 8 Mar 2021 18:41:46 -0500 Subject: [PATCH] math: add more probability rules and slight refactor --- docs/mcv4u7.md | 15 ++++++++++++--- 1 file changed, 12 insertions(+), 3 deletions(-) diff --git a/docs/mcv4u7.md b/docs/mcv4u7.md index 51005f0..fc65772 100644 --- a/docs/mcv4u7.md +++ b/docs/mcv4u7.md @@ -174,11 +174,20 @@ $$P(A)=\frac{n(A)}{n(U)}$$ As event $A$ must be a subset of all possible outcomes $U$, where $1$ indicates that the event always happens and $0$ the opposite: $$0\leq P(A)\leq 1$$ -The complement of **event A** is the probability that it does not happen. It is written as $A^c$, $A'$, or $\pu{not } A$. +The **complement** of event $A$ is the probability that it does not happen. It is written as $A^c$, $A'$, or $\pu{not } A$. $$P(A')=1-P(A)$$ -Events $A$ and $B$ are disjoint if no outcomes between them are common and can never happen simultaneously. As such the probability of one of the events happening is equal to their sum. -$$P(A\cup B)=P(A)+P(B)$$ +Events $A$ and $B$ are **disjoint** or mutually exclusive if no outcomes between them are common and can never happen simultaneously. As such the probability of one of the events happening is equal to their sum. +$$ +P(A\cup B)=P(A)+P(B) \\ +P(A\cap B)=0 +$$ + +Events $A$ and $B$ are **exhaustive** if their union includes all possible outcomes in the sample space: $A\cup B=U$. +$$P(A\cup B)=1$$ + +The **principle of inclusion and exclusion** forms a general rule for the union between two sets: +$$P(A\cup B)=P(A)+P(B)-P(A\cap B)$$ ## Resources