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03. Conditional Probability

$\gdef \N{\mathbb{N}} \gdef \Z{\mathbb{Z}} \gdef \Q{\mathbb{Q}} \gdef \R{\mathbb{R}} \gdef \C{\mathbb{C}} \gdef \setcomp#1{\overline{#1}} \gdef \sseq{\subseteq} \gdef \pset#1{\mathcal{P}(#1)} \gdef \covariant{\operatorname{Cov}} \gdef \of{\circ} \gdef \p{^{\prime}} \gdef \pp{^{\prime\prime}} \gdef \ppp{^{\prime\prime\prime}} \gdef \pn#1{^{\prime\times{#1}}} $

Definition

Conditional Probability is the probability of A if B occurs. Denoted by \(P(A|B)\).

A Die Is Rolled

Let A be the event of rolling greater than a three.
Let B be the event of rolling an even number.
\(\Omega=\{1,2,3,4,5,6\}\)
We are told that B happens. Now what is the probability of A?

\(\Omega_{\text{new}}=\{2,4,6\}\)
Given our \(\Omega_{\text{new}}\), \(A_{\text{new}}=\{4,6\}\)

\[\frac{|A_{\text{new}}|}{|\Omega_{\text{new}}|}=\frac23\]

The Formula

\[P(A|B)=\frac{P(A\cap B)}{P(B)}\]

Reuven Has 2 Children

Let A be the event that Reuven has exactly 2 boys.
Let B be the event that Reuven has at least one boy.
Given that Reuven has at least one boy, what is the probability of Reuven having 2 boys?

Applying The Formula, \(P(A|B)=\frac{P(A\cap B)}{P(B)}\)
Since \(A\subseteq B\), \(\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}=\frac{\frac14}{\frac34}=\frac13\)

\[\frac{P(A\cap B)}{P(B)}=\frac13\]

Independent Events

Events \(A\) and \(B\) are independent events  if \(P(A|B)=P(A)\)
(Meaning, if knowing \(B\) happened has no effect on the probability of \(A\).)

Properties of Independent Events
  1. Independence is mutual.
    • If \(P(A|B)=P(A)\), then \(P(B|A)=P(B)\).
  2. \(A, B\) are independent \(\iff P(A\cap B)=P(A)\cdot P(B)\)
    • This extends to \(n\) mutually independent events.

Don’t mix this up with disjoint events!

Disjoint events are not the same thing as independent events.

In fact, if \(P(A),P(B)\not=0\), then if they are disjoint \(A,B\) must be dependent.

Evaluating Conditional Probabilities

Example

\(A=\) I visit New York
\(B=\) I visit the Statue of Liberty
Excalidraw Excalidraw
Given \(P(A)=0.7\) and \(P(B|A)=0.4\), what is \(P(B)\)?

\(P(B)=P(B\cap A)\) because \(B\subseteq A\).
By definition: \(P(B|A)=\frac{P(B\cap A)}{P(A)}\).
Multiplying both sides by \(P(A)\) gives us:

\[\begin{align}P(B\cap A)&=P(A)\cdot P(B|A)\\P(B\cap A)&=0.7\cdot 0.4&=0.28\\P(B\cap A)&=P(B)&=0.28\end{align}\]

Law of Total Probability (LTP)

Introductory Example

Given 2 jars

Excalidraw Excalidraw

The probability of picking jar 1 is \(\frac14\), and picking jar 2 is \(\frac34\). A ball is selected at random from the chosen jar. What is the probability that I select a blue ball?

Define events:

\[\begin{align}A_1&=\text{Choose jar }1\\A_2&=\text{Choose jar }2\\B&=\text{Choose a blue ball}\\\end{align}\]

We know \(P(A_1)=\frac14\), \(P(A_2)=\frac34\).
\(P(B\mid A_1)=\frac23\), \(P(B\mid A_2)=\frac13\).
So, how do we find \(P(B)\)?

Note that \(B=(B\cap A_1)\cup(B\cap A_2)\), which are disjoint events.
\(B\cap A_1\subseteq A_1\), hence, \(P(B\cap A_1)=P(A_1)\cdot P(B\mid A_1)\)
\(B\cap A_2\subseteq A_2\), hence, \(P(B\cap A_2)=P(A_2)\cdot P(B\mid A_2)\)

\[\begin{align}P(B)&=P(B\cap A_1)&+&P(B\cap A_2)\\&=P(A_1)\cdot P(B\mid A_1)&+&P(A_2)\cdot P(B\mid A_2)\\&=\frac14\cdot\frac23&+&\frac34\cdot\frac13\\\\&=\frac5{12}\end{align}\]

Let \(A_1,\dots A_n\) be disjoint events which partition \(\Omega\).
Let \(B\) be any event.
Then:

The Law

\(B\) is the union of the intersection of \(A_k\) and \(B\), for all \(k\): 
\(B=(B\cap A_1)\cup(B\cap A_2)\cup\dots\cup(B\cap A_n)\)
Then,

\[P(B)=\sum_{k=1}^n{P(A_k)\cdot P(B|A_k)}\]
Proof

This is very likely on the ProbabilityExam{: #ProbabilityExam .hash}

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