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:

• $\mathbb P(B^c)$
$\mathbb P(B^c)=1-\mathbb P(B)\\ =1-\frac13\\ =\frac23$
• $\mathbb P(A^c\cap B)$
$\mathbb P(B)=\mathbb P(A\cap B)+\mathbb P(A^c\cap B)\\ A\cap A^c=\emptyset,\quad A\cup A^c=\Omega\\ \mathbb P(A^c\cap B)=\mathbb P(B)-\mathbb P(A\cap B)\\ =\frac 7{30}$
• $\mathbb P(A^c\cup B)$
$\mathbb P(A^c\cup B)\\ =\mathbb P(A^c)+\mathbb P(B)-\mathbb P(A^c\cap B)\\ =1-\mathbb P(A)+\mathbb P(B)-\mathbb P(A^c\cap B)\\ =1-\frac13+\frac13-\frac 7{30}$

### 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

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