ELEC 321

Tutorial 1

Updated 2017-09-15

Review

Properties

  1. \[\mathbb P(\emptyset)=0\]
  2. \[\mathbb P(A^c)=1-\mathbb P(A)\]
  3. \[A\subset B\implies\mathbb P(A)\leq\mathbb P(B)\]

Union Bound (Boole’s Inequality)

\[\mathbb P(\bigcup^n_{i=1}A-i)\leq\sum^n_{i=1}\mathbb P(A_i)\]

Total Law of Probability

Given \(E_1, E_2,\dotsc,E_n\),

Condition 1: \(E_i\cap E_j = \emptyset; i\neq j\)

Condition 2 (exhaustive): \(\bigcup^n_{i=1}E_i=\Omega\), so \(\mathbb P(A)=\sum^n_{i=1}\mathbb P(A\cap E_i)\)

Conditional Probability

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

Total Law of Property expressed in conditional property:

\[\mathbb P(A)=\sum^n_{i=1}\mathbb P(E_i)\mathbb P(A\vert E_i)\]

Example: Binary Communication System

There is a \(\frac23\) probability of sending a 0, and \(\frac13\) probability of sending a 1. If the bit sent is a 0, there is 0.9 that the output is correct, 0.1 probability of flipped bit. Else if the bit sent is a 1, there is 0.8 chance that that output is correct, but 0.2 chance that the bit is flipped.

Let

\[X: \text{input}\\Y:\text{output}\]

So

\[\mathbb P(X=0)=\frac23\\ \mathbb P(X=1)=\frac13\\ \mathbb P(Y=0\vert X=0)=0.9\\ \mathbb P(Y=1\vert X=0)=0.1\\ \mathbb P(Y=0\vert X=1)=0.2\\ \mathbb P(Y=1\vert X=1)=0.8\\\]

To find the probability of error, we add the total probability of each error up (total law property)

\[\mathbb P(Y\neq X)=\mathbb P(Y\neq X\vert X=0)\mathbb P(X=0)+\mathbb P(Y\neq X\vert X=1)\mathbb P(X=1)\\ =\mathbb P(Y=1\vert X=0)\mathbb P(X=0)+\mathbb P(Y=0\vert X=1)\mathbb P(X=1)\\ =(0.1)(\frac23)+(0.2)(\frac13)\\\]

Practice

A1

Given \(\mathbb P(A)=\frac13, \mathbb P(B)=\frac13,\mathbb P(A\cap B)=\frac 1{10}\), find:

A6

Family has two children, and

\[B=\{\text{one of the children is s boy}\}\\ A=\{\text{both children are boys}\}\\\]

We want to find \(\mathbb P(A\vert B)\)

\[\Omega=\{(B,B), (B,G), (G,B), (G,G)\}\]

Since they are all equally likely, the probability of each one happening is \(\frac14\)

\[A=\{(B,B), (B,G), (G,B)\}\] \[B=\{(B,B)\}\]

And we can know \(A\cap B=\{(B,B)\}\) and find the conditional probability.

\[\mathbb P(A\vert B)=\frac{\mathbb P(A\cap B)}{\mathbb P(B)}\\ =\frac{0.25}{\mathbb P(\{(B,B), (B,G), (G,B)\})}\\ =\frac{0.25}{0.75}\\ =\frac13\] \[\mathbb P(A\vert B)=\frac{\mathbb P(A\cap B)}{\mathbb P(B)}\\ =\frac{0.25}{\mathbb P(\{(B,B), (B,G), (G,B)\})}\\ =\frac{0.25}{0.75}\\ =\frac13\]

A7

There are \(n\) people in the room, we want the probability of at least 2 people having same birthday.

Let

\[A=\{\text{at least 2 people having same birthday}\}\\ A^c=\{\text{none of the people having the same birthday}\}\] \[\mathbb P(A)=1-\mathbb P(A)\]

The total possibilities of date of birth is \(365\) days. and \(365-(n-1)\) is number of possibilities with constrain that no people share birthdays.

And

\[\mathbb P(A^c)=(365)(365-1)(365-2)(365-3)\dots(365-(n-1))=\frac{365!}{365^n(365-n)!}\]

So

\[\mathbb P(A)=1-\frac{365!}{365^n(365-n)!}\]