Chapter 2: Conditional Probability  
MATH/STAT 394: Probability I  
Arman Jahangiri  
University of Washington Department of Mathematics  
Summer 2026  
Arman Jahangiri  
Chapter 2: Conditional Probability  
Summer 2026  
1
 
Learning Goals  
By the end of this lecture, students should be able to:  
▶ interpret conditional probability intuitively;  
▶ apply Bayes’ rule and the Law of Total Probability;  
▶ distinguish between P(A|B) and P(B|A);  
▶ understand independence and conditional independence;  
▶ use conditioning as a strategy in solving in probability problems.  
Arman Jahangiri  
Chapter 2: Conditional Probability  
Summer 2026  
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Conditional Probability  
Arman Jahangiri  
Chapter 2: Conditional Probability  
Summer 2026  
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Why Do We Need Conditional Probability?  
In many situations, probabilities change once additional information becomes available.  
Before observing new information  
After observing new information  
Let  
Suppose we now observe dark clouds:  
R = {it rains today}.  
Without any additional information, we assign the  
C = {dark clouds are present}.  
probability  
The probability of rain may now change to  
P(R).  
P(R | C),  
This represents our initial uncertainty about rain.  
read as “probability of R given C.”.  
An important observation here is that we do not necessarily have P(R | C) = P(R), because the information that  
C occurred changes what outcomes now appear more plausible.  
Main Idea  
Conditional probability quantifies how probabilities are updated after new information or evidence is observed.  
Arman Jahangiri  
Chapter 2: Conditional Probability  
Summer 2026  
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Definition of Conditional Probability  
Definition  
If  
P(B) > 0,  
then the conditional probability of A given B is  
P(A ∩ B)  
P(A|B) =  
.
P(B)  
Interpretation:  
▶ B is the observed evidence;  
▶ we restrict attention to outcomes in B;  
▶ probabilities are renormalized within B.  
Arman Jahangiri  
Chapter 2: Conditional Probability  
Summer 2026  
5
Visual Interpretation of Conditional Probability  
When conditioning on B, the event B becomes the new sample space (i.e., replaces S). The conditional probability P(A | B)  
measures the proportion of B that also belongs to A.  
P(A ∩ B)  
P(A | B) =  
Image adapted from: ExcelR — Conditional Probability.  
provided P(B) > 0.  
P(B)  
Arman Jahangiri  
Chapter 2: Conditional Probability  
Summer 2026  
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Example: Two Cards  
Two cards are drawn without replacement.  
Define:  
A = {first card is a heart}, B = {second card is red}.  
Then,  
13 · 25  
25  
P(A ∩ B) =  
=
.
52 · 51  
204  
Also,  
1
1
P(B) =  
= 0.5,  
P(A) =  
= 0.25.  
2
4
Therefore,  
25/204  
25  
P(A|B) =  
=
≈ 0.245 < P(A) = 0.25,  
1/2  
102  
while  
25/204  
25  
P(B|A) =  
=
≈ 0.49 < P(B) = 0.5.  
51  
1/4  
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Chapter 2: Conditional Probability  
Summer 2026  
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Important Warning  
Conditional probabilities are directional  
In general,  
P(A|B) = P(B|A).  
̸
The conditioning event matters.  
P(A|B)  
means:  
probability of A after learning B.  
Arman Jahangiri  
Chapter 2: Conditional Probability  
Summer 2026  
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Bayes’ Rule and LOTP  
Arman Jahangiri  
Chapter 2: Conditional Probability  
Summer 2026  
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Intersection Formula  
From the definition:  
P(A ∩ B)  
P(A|B) =  
.
P(B)  
Multiplying both sides by P(B) gives:  
P(A ∩ B) = P(B)P(A|B).  
Similarly,  
P(A ∩ B) = P(A)P(B|A).  
The last two equalities together imply an important equality in Probability, known as the Bayes’ rule, given in the  
next slide.  
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Chapter 2: Conditional Probability  
Summer 2026 10  
Bayes’ Rule: Updating Beliefs with Evidence  
Bayes’ Rule  
P(B | A)P(A)  
P(A | B) =  
.
P(B)  
Bayes’ rule describes how probabilities ”evolve” when new information is observed.  
P(B | A)  
P(A)  
before observing B  
=⇒  
P(A | B) = P(A) ×  
after observing B  
|{z}  
|
{z  
}
P(B)  
updating weight  
|
{z  
}
The observation of B changes our belief about whether A occurred.  
Interpretation  
Bayes’ rule updates a prior belief P(A) into a posterior belief P(A | B) by multiplying by a factor that  
measures how strongly the evidence B supports A.  
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Chapter 2: Conditional Probability  
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Law of Total Probability  
Suppose  
A1, . . . , An  
form a partition of S.  
Then:  
n
X
P(B) =  
P(B|Ai )P(Ai ).  
i=1  
Interpretation  
Break the sample space into cases, compute probabilities within each case, then combine them.  
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Chapter 2: Conditional Probability  
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Rare Disease Example  
Suppose we are analyzing the disease rate in a certain area.  
Let  
D = {person has the disease}, T = {test result is positive},  
and assume  
P(D) = 0.01, P(T | D) = 0.95, P(Tc | Dc ) = 0.95.  
Interpretation:  
▶ The disease prevalence is 1%;  
▶ The test correctly detects the disease 95% of the time;  
▶ The test correctly returns a negative result 95% of the time for healthy individuals.  
Question: A patient tests positive. What is P(D | T)?  
Using Bayes’ Rule and LOTP:  
Using Bayes’ Rule and the LOTP,  
P(T | D)P(D)  
P(T | D)P(D)  
0.95(0.01)  
P(D | T) =  
Arman Jahangiri  
=
=
≈ 0.16.  
0.95(0.01) + 0.05(0.99)  
P(T)  
P(T | D)P(D) + P(T | Dc )P(Dc )  
Chapter 2: Conditional Probability  
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Why This Feels Surprising  
Most people expect the answer to be close to 95%.  
But the disease is very rare.  
Among many healthy people:  
5%  
still test positive.  
Those false positives greatly outnumber the true positives.  
Lesson  
Posterior probabilities depend both on:  
▶ the evidence,  
▶ and the prior probability.  
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Conditional Probability as Probability  
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Conditional Probabilities Behave Like Probabilities  
Fix an event E with  
P(E) > 0.  
Then:  
P(·|E)  
satisfies all probability axioms.  
In particular:  
0 ≤ P(A|E) ≤ 1,  
P(S|E) = 1,  
P(Ac |E) = 1 − P(A|E).  
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Chapter 2: Conditional Probability  
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Independence  
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Chapter 2: Conditional Probability  
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Definition of Independence  
Definition (Independence)  
Events A and B are independent if  
P(A | B) = P(A),  
provided P(B) > 0.  
Interpretation:  
Learning that B occurred gives no information about A (and vice versa⋆).  
(⋆) We will later show that  
P(A | B) = P(A) ⇐⇒ P(B | A) = P(B).  
Thus independence is symmetric:  
A is independent of B ⇐⇒ B is independent of A.  
Think of it this way: P(A with new (additional) information) = P(A with no additional information)  
Lemma (Equivalent characterization of independence)  
Events A and B are independent if and only if  
P(A ∩ B) = P(A) × P(B).  
Arman Jahangiri  
Chapter 2: Conditional Probability  
Summer 2026 18  
Conditional Independence  
Conditional Independence  
Events A and B are conditionally independent given E if  
P(A ∩ B|E) = P(A|E)P(B|E).  
Conditional independence is different from ordinary independence.  
Two events may:  
▶ be independent but not conditionally independent;  
▶ be conditionally independent but not independent.  
Arman Jahangiri  
Chapter 2: Conditional Probability  
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Example: Hidden Coin Type  
Suppose:  
▶ with probability 1/2 we choose a fair coin;  
▶ with probability 1/2 we choose a biased coin.  
Conditional on which coin was chosen:  
coin tosses are independent.  
But unconditionally:  
coin tosses are dependent,  
because observing earlier tosses gives information about which coin was chosen.  
Arman Jahangiri  
Chapter 2: Conditional Probability  
Summer 2026 20  
Conditioning as a Strategy  
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Chapter 2: Conditional Probability  
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Monty Hall Problem: Setup  
There are three doors:  
▶ one door hides a car;  
▶ two doors hide goats.  
You choose one door, but it is not opened yet.  
Assume, without loss of generality, that you initially choose door  
1.  
Ci = {car is behind door i}.  
Before Monty opens a door,  
Image source: Brilliant: Monty Hall Problem.  
1
P(C1) = P(C2) = P(C3) =  
.
3
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Chapter 2: Conditional Probability  
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Monty Hall Problem: Why Switching Wins  
Monty knows where the car is and always opens a door with a goat.  
So the important question is:  
Should you switch?  
Car location First choice Switching  
C1  
C2  
C3  
correct  
wrong  
wrong  
loses  
wins  
wins  
Therefore,  
1
3
1
3
2
3
Image source: Brilliant: Monty Hall Problem.  
P(win by switching) = P(C2 ∪ C3) = P(C2) + P(C3) =  
+
=
.
Arman Jahangiri  
Chapter 2: Conditional Probability  
Summer 2026 23  
Monty Hall Problem via Bayes’ Rule  
Suppose Monty opens door 3.  
Define  
M3 = {Monty opens door 3}.  
Thus:  
1
2
3
P(initial choice correct) =  
door 1  
,
and  
P(initial choice wrong) =  
door 2 or 3  
.
|
{z  
}
3
|
{z  
}
After Monty reveals a goat behind door 3, the entire remaining probability mass concentrates on door 2.  
Using Bayes’ Rule,  
P(M3 | C2)P(C2)  
P(win by switching|M3) = P(C2 | M3) =  
.
P(M3)  
Now, P(C2) = 1 , P(M3 | C2) = 1, and  
3
1
2
1
3
1
1
3
1
P(M3) =  
·
+ 1 · + 0 ·  
=
.
3
2
Therefore,  
1
3
1 ·  
2
3
1
P(C2 | M3) =  
=
, and thus , and thus P(win without switching|M3) = P(C1|M3) =  
1
3
Arman Jahangiri  
Chapter 2: Conditional Probability  
Summer 2026 24  
Summary  
▶ Conditional probability updates probabilities using observed evidence.  
▶
P(A ∩ B)  
P(A|B) =  
.
P(B)  
▶ Bayes’ Rule relates:  
P(A|B) and P(B|A).  
▶ LOTP breaks problems into cases.  
▶ Conditional probabilities themselves satisfy the probability axioms.  
▶ Independence means:  
P(A|B) = P(A).  
▶ Conditioning is a powerful problem-solving strategy.  
Next lecture  
Random variables and distributions.  
Arman Jahangiri  
Chapter 2: Conditional Probability  
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