Lecture 5: conditional probability

Reading: Cameron chapter 2
Last semester's notes
Definitions: conditional probability
Modeling with conditional probability
Bayes's rule, Law of total probability

Conditional probability

Definition: If $A$ and $B$ are events, then the probability of A given B, written $Pr(A|B)$ is given by $Pr(A|B) := \frac{Pr(A \cap B)}{Pr(B)}$ Note that $Pr(A|B)$ is only defined if $Pr(B) \neq 0$ .

Intuitively, $Pr(A|B)$ is the probability of $A$ in a new sample space created by restricting our attention to the subset of the sample space where $B$ occurs. We divide by $Pr(B)$ so that $Pr(B|B) = 1$ .

Note: $A|B$ is not defined, only $Pr(A|B)$ ; this is an abuse of notation, but is standard.

Probability trees

Using conditional probability, we can draw a tree to help discover the probabilities of various events. Each branch of the tree partitions part of the sample space into smaller parts.

For example: suppose that it rains with probability 30%. Suppose that when it rains, I bring my umbrella 3/4 of the time, while if it is not raining, I bring my umbrella with probability 1/10. Given that I bring my umbrella, what is the probability that it is raining?

One way to model this problem is with the sample space

$S = \{raining (r), not raining (nr)\} \times \{umbrella (u), no umbrella (nu)\} = \{(r,u), (nr,u), (r,nu), (nr,nu)\}$

Let $R$ be the event "it is raining". Then $R = \{(r,u), (r,nu)\}$ . Let $U$ be the event "I bring my umbrella". Then $U = \{(r,u), (nr,u)\}$ .

The problem tells us that $Pr(R) = 3/10$ . It also states that $Pr(U|R) = 3/4$ while $Pr(U|\bar{R}) = 1/10$ . We can use the following fact:

Fact: $Pr(\bar{A}|B) = 1 - Pr(A|B)$ . Proof left as exercise.

to conclude that $Pr(\bar{U}|R) = 1/4$ and $Pr(\bar{U}|\bar{R}) = 9/10$ .

We can draw a tree:

Probability tree (LaTeX source)

We can compute the probabilities of the events at the leaves by multiplying along the paths. For example, $Pr(\{(r,u)\}) = Pr(U \cap R) = Pr(R)Pr(U | R) = (3/10)(3/4) = (9/40)$

To answer our question, we are interested in $Pr(R|U) = Pr(U \cap R)/Pr(U)$ . We know $U = \{(u,r), (u,nr)\}$ . We can compute $Pr(U)$ using the third axiom; $Pr(U) = Pr(\{(u,r)\}) + Pr(\{(u,nr)\}) = (3/10)(3/4) + (7/10)(1/10)$ . We can then plug this in to the above formula to find $Pr(R|U)$ .

Note we could also answer this using Bayes's rule and the law of total probability (see below); it would amount to exactly the same calculation. The tree just helps organize all of the variables.

Bayes's Rule

Bayes's rule is a simple way to compute $P(A|B)$ from $P(B|A)$ .

Claim: (Bayes's rule): $P(A|B) = P(B|A)P(A)/P(B)$ .

Proof: Write down the definitions; proof left as exercise.

Law of total probability

Claim: (law of total probability) If $A_1, \dots, A_n$ partition the sample space $S$ (that is, if $A_i \cap A_j = \emptyset$ for $i \neq j$ and $S = \cup A_i$ ), then

$Pr(B) = \sum_{i} Pr(B|A_i)Pr(A_i)$

Proof sketch: Write $B = \cup_{i} (B \cap A_i)$ . Apply third axiom to conclude $Pr(B) = \sum_{i} Pr(B \cap A_i)$ . Apply definition of $Pr(B | A_i)$ .