Lecture 13: conditional probability


Definition: Two events \(A\) and \(B\) are independent if \(Pr(A \cap B) = Pr(A)Pr(B)\).

Important: Do not assume events are independent unless given a good reason to do so.

Example: Suppose I roll two 6-sided dice. For either die, the probability of getting any number from 1...6 is 1/6. What is the probability of getting a pair of twos?

Answer: it could be anything from 0 to 1/6. The dice could be taped together in a way that it is impossible to get (2,2) while still making the probability of any given roll for either die to be 1/6.

If we are also told that the two rolls are independent, then we can conclude that \[\begin{aligned} Pr(\{(2,2)\}) &= Pr(\{(2,n)\mid n \in \{1,\dots,6\}\} \cap \{(n,2) \mid n \in \{1,\dots,6\}\}) \\ &= Pr(\{(2,n)\mid n \in \{1,\dots,6\}\})\cdot Pr(\{(n,2) \mid n \in \{1,\dots,6\}\}) \\ &= (1/6)(1/6) \\ \end{aligned} \]

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.

Alternative definition of independence

Conditional probability can be used to give an equivalent definition of independence:

Claim: \(A\) and \(B\) are independent if and only if \(Pr(A|B) = Pr(A)\).

This is perhaps a more intuitive notion of independence: if I tell you that \(B\) happened, it doesn't change your estimate of the probability that \(A\) happens.

Proof: (⇒) Suppose \(A\) and \(B\) are independent. We wish to show \(Pr(A|B) = Pr(A)\). Well, by definition, \(Pr(A|B) = Pr(A \cap B)/Pr(B)\). Since \(A\) and \(B\) are independent, \(Pr(A \cap B) = Pr(A)Pr(B)\). Plugging this in, we see that \(Pr(A|B) = Pr(A)Pr(B)/Pr(B) = Pr(A)\).

(⇐) Suppose \(Pr(A|B) = Pr(A)\). Then \(Pr(A) = Pr(A \cap B)/Pr(B)\). Multiplying both sides by \(Pr(B)\) gives the desired result.

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: left as exercise.

Example: See next lecture

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)

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 \(Pr(R|U) = Pr(U \cap R)/Pr(U) = (9/40)/(3/10) = 3/4\).

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