Conditional Probability, Independence of Events

Class: STAT 131 Subject: probability Date: 2025-01-21 Teacher: Prof. Marcela

Conditional Probability

• given $A$ and $B$ are events with $P(B)>0$, the the conditional probability of $A$ given $B$, denoted by $P(A|B)$, is defined as: $P(A|B)=\frac{P(A\cap B)}{P(B)}$

Example

• Standard deck: 52 cards that are shuffled.
• 2 cards are drawn randomly, one at a time without replacement.
• Let $A$ be the event that the first card is a heart.
• Let $B$ be the event that the second card is red.
• Find $P(A|B)$ and $P(B|A)$. Are they equal?

Using the naive definition of probability and the multiplication rule: $P(A\cap B)=\frac{13\ast 25}{52\ast 51}=\frac{25}{204}$ note: we use 25/51 because the first card would be red resulting in one less red card in the deck. $P(A)=\frac{1}{4}$ and $P(B)=\frac{1}{2}$

$P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{25/204}{1/2}=\frac{25}{102}$

Bayes’ Rule

• Probability of the intersection of two events. For any events $A$ and $B$ with positive probabilities: $P(A\cap B)=P(B)P(A|B)=P(A)P(B|A)$

Independence of Events

• Events $A$ and $B$ are independent if $P(A\cap B)=P(A)P(B)$
• If $P(A)>0$ and $P(B)>0$, then this is equivalent to: $P(A|B)=P(A)$
• which is also equivalent to: $P(B|A)=P(B)$
• Basically, if non of the events have no effect on each other, than they are independent

Proposition

• If $A$ and $B$ are independent, then $A$ and $B^c$ are independent, $A^c$ and $B$ are independent, and $A^c$ and $B^c$ are independent. $P(B^c|A)=1-P(B|A)=1-P(B)=P(B^c)$

$P(D_1\mid W)=\frac{P(D_1\cap W)}{P(W)}$

$w_o\ast P(D_1^c)\ast P(D_2^c)+1\ast P(D_1\cup D_2)$ $=$ $w_o\ast (1-p_1)(1-p_2)+p_1+p_2-p_1\ast p_2$ $=$ $w_o\ast (1-p_1)(1-p_2)+P(D_1)+P(D_2)-P(D_1\cap D_2)$ $P(bothgirls\cap atleastonec)=P(atleastoneC\mid bothgirls)\ast P(bothgirls)$ $=$ $1-P(noC\mid bothgirls)=1-(1-p)(1-p)=1-(1-p{)}^2$ $=$ $0.25\ast (1-(1-p{)}^2)=0.25\ast p(2-p)$ $P(D_1\mid W)P(D_2\mid W)=\frac{p_1}{P(W)}\ast \frac{p_2}{P(W)}\ast \frac{p_1{p}_2}{P(W)}$ Because $P(D_1\cap D_2\mid W)=P(D_1\mid W)P(D_2\mid W)$ ,  D_1  and  D_2  are conditionally independent given  W

$P(W)=P(W\mid D_1\cup D_2)P(D_1\cup D_2)=1\cdot (p_1+p_2-p_1{p}_2)$ Because $P(D_1\cap D_2\mid W)=P(D_1\mid W)P(D_2\mid W)$, conditional independence is still observed