The Central Idea
Suppose we want the probability of event \(A\), but we already know that event \(B\) has occurred.
Outcomes outside \(B\) are no longer relevant.
The effective sample space becomes \(B\), and the favorable outcomes are those that lie in both \(A\) and \(B\).
\[ \boxed{ \text{conditioning on }B = \text{restricting attention to }B }. \]Conditional Probability Notation
The conditional probability of \(A\) given \(B\) is written
\[ P(A\mid B). \]It is read:
“the probability of \(A\) given \(B\).”
The vertical bar
\[ \mid \]means “given that.”
The Conditional Probability Formula
If
\[ P(B)>0, \]then
\[ \boxed{ P(A\mid B) = \frac{P(A\cap B)}{P(B)} }. \]The numerator measures the outcomes satisfying both conditions. The denominator rescales probability to the event \(B\), which now acts as the effective sample space.
Why \(P(B)\) Must Be Positive
The formula divides by
\[ P(B). \]Therefore, the elementary definition requires
\[ P(B)>0. \]Conditioning on an event with probability zero requires more advanced probability theory and is outside the scope of this lesson.
Conditioning Changes the Denominator
Unconditional probability compares an event with the full sample space:
\[ P(A). \]Conditional probability compares the overlap \(A\cap B\) with \(B\):
\[ P(A\mid B) = \frac{P(A\cap B)}{P(B)}. \]The essential change is the denominator.
Example: One Die
Roll a fair six-sided die.
Let
\[ A=\{\text{result is greater than }3\} = \{4,5,6\} \]and
\[ B=\{\text{result is even}\} = \{2,4,6\}. \]Given that \(B\) occurred, the effective sample space is
\[ \{2,4,6\}. \]The outcomes also satisfying \(A\) are
\[ \{4,6\}. \]Therefore,
\[ \boxed{ P(A\mid B)=\frac23 }. \]The Same Example Using the Formula
For the fair die,
\[ P(B)=\frac36=\frac12, \]while
\[ A\cap B=\{4,6\} \]gives
\[ P(A\cap B)=\frac26=\frac13. \]Thus,
\[ P(A\mid B) = \frac{1/3}{1/2} = \boxed{\frac23}. \]Conditional Probability by Counting
In a finite equally likely sample space,
\[ P(A\mid B) = \frac{|A\cap B|}{|B|}. \]This follows because
\[ P(A\cap B)=\frac{|A\cap B|}{|S|} \]and
\[ P(B)=\frac{|B|}{|S|}. \]Dividing cancels \(|S|\).
Example: Conditional Integer Selection
One integer is selected uniformly from
\[ \{1,2,\ldots,30\}. \]Given that the number is even, find the probability that it is divisible by \(6\).
There are
\[ 15 \]even integers.
The multiples of \(6\) are
\[ 6,12,18,24,30, \]so there are \(5\) favorable outcomes.
Therefore,
\[ \boxed{ P(\text{divisible by }6\mid\text{even}) = \frac5{15} = \frac13 }. \]Order Matters in Conditional Notation
In general,
\[ P(A\mid B) \]and
\[ P(B\mid A) \]are different.
The event after the vertical bar determines the restricted sample space.
Example: Reversing the Condition
Using the previous die events,
\[ A=\{4,5,6\} \]and
\[ B=\{2,4,6\}. \]We found
\[ P(A\mid B)=\frac23. \]But conditioning on \(A\) gives the effective sample space
\[ \{4,5,6\}. \]Within that set, the even outcomes are
\[ \{4,6\}, \]so
\[ \boxed{ P(B\mid A)=\frac23 }. \]These happen to match in this example, but that equality is not automatic.
Example Where the Directions Differ
Suppose
\[ P(A)=0.50, \qquad P(B)=0.20, \qquad P(A\cap B)=0.10. \]Then
\[ P(A\mid B) = \frac{0.10}{0.20} = \boxed{0.50}, \]while
\[ P(B\mid A) = \frac{0.10}{0.50} = \boxed{0.20}. \]Reversing the conditioning changes the denominator.
Conditional Probability and the Multiplication Rule
Starting with
\[ P(A\mid B) = \frac{P(A\cap B)}{P(B)}, \]multiply both sides by \(P(B)\):
\[ \boxed{ P(A\cap B) = P(B)P(A\mid B) }. \]Similarly,
\[ \boxed{ P(A\cap B) = P(A)P(B\mid A) }. \]Two Views of the Same Relationship
Conditional probability can be solved for from an intersection:
\[ P(A\mid B) = \frac{P(A\cap B)}{P(B)}. \]Or an intersection can be constructed from a conditional probability:
\[ P(A\cap B) = P(B)P(A\mid B). \]These are rearrangements of the same relationship.
Example: Sequential Draws
Two cards are drawn from a standard deck without replacement.
Let \(A\) be “the first card is an ace” and \(B\) be “the second card is an ace.”
Then
\[ P(A)=\frac4{52}. \]Given that the first card was an ace,
\[ P(B\mid A)=\frac3{51}. \]Therefore,
\[ P(A\cap B) = \frac4{52}\cdot\frac3{51} = \boxed{\frac1{221}}. \]Why Without-Replacement Probabilities Change
Conditioning reflects the information created by the first draw.
If the first card is known to be an ace, the deck now contains
\[ 51 \]cards, of which
\[ 3 \]are aces.
The conditional probability
\[ \frac3{51} \]reflects the updated sample space.
With Replacement Can Remove the Dependence
Suppose the first card is replaced and the deck is reshuffled before the second draw.
Then even after learning that the first card was an ace,
\[ P(\text{second ace}\mid\text{first ace}) = \frac4{52}. \]The original composition has been restored.
Conditional Probability in Two-Way Tables
Two-way tables organize observations according to two categorical variables.
Conditional probability restricts attention to one row, one column, or another specified category.
The denominator becomes the total count in the conditioning category.
Example: Two-Way Table
Suppose a group contains \(100\) people classified as follows:
| Uses Product | Does Not Use Product | Total | |
|---|---|---|---|
| Group A | 30 | 20 | 50 |
| Group B | 10 | 40 | 50 |
| Total | 40 | 60 | 100 |
If a randomly selected person is known to be in Group A, the relevant denominator is \(50\), not \(100\).
Conditional Probability from the Table
Let \(U\) mean “uses the product” and \(A\) mean “belongs to Group A.”
The table gives
\[ |U\cap A|=30 \]and
\[ |A|=50. \]Therefore,
\[ \boxed{ P(U\mid A)=\frac{30}{50}=0.60 }. \]Reverse the Conditioning
Now ask for the probability that a product user belongs to Group A:
\[ P(A\mid U). \]There are \(40\) product users, of whom \(30\) are in Group A.
Therefore,
\[ \boxed{ P(A\mid U)=\frac{30}{40}=0.75 }. \]Again,
\[ P(U\mid A)\neq P(A\mid U). \]Rows and Columns Become Conditional Sample Spaces
In a two-way table:
- conditioning on a row uses that row total as the denominator;
- conditioning on a column uses that column total as the denominator.
The conditioning information determines which portion of the table remains relevant.
Tree Diagrams and Conditional Probability
Probability trees display conditional probabilities naturally.
A first branch may represent event \(A\) or \(A^c\).
From the \(A\) branch, later branches represent probabilities such as
\[ P(B\mid A) \]and
\[ P(B^c\mid A). \]The branch probabilities from the same node sum to \(1\).
Multiply Along a Tree Path
A completed path corresponds to an intersection of events.
For the path \(A\) followed by \(B\),
\[ \boxed{ P(A\cap B)=P(A)P(B\mid A) }. \]Thus, tree diagrams visually encode the multiplication rule.
Add Across Disjoint Tree Paths
If an event can occur along several mutually exclusive completed paths, add their probabilities.
For example, \(B\) can occur together with either \(A\) or \(A^c\):
\[ B=(A\cap B)\cup(A^c\cap B). \]The two pieces are disjoint, so
\[ \boxed{ P(B) = P(A)P(B\mid A) + P(A^c)P(B\mid A^c) }. \]The Law of Total Probability in a Two-Part Partition
Since \(A\) and \(A^c\) partition the sample space,
\[ \boxed{ P(B) = P(B\mid A)P(A) + P(B\mid A^c)P(A^c) }. \]This formula combines conditional probabilities across all possible conditioning branches.
General Law of Total Probability
Suppose
\[ A_1,A_2,\ldots,A_n \]form a partition of the sample space and each has positive probability.
Then for an event \(B\),
\[ \boxed{ P(B) = \sum_{i=1}^{n} P(B\mid A_i)P(A_i) }. \]The event \(B\) is decomposed according to the mutually exclusive ways the partition can occur.
Example: Manufacturing Sources
Suppose \(60\%\) of items come from machine \(A\) and \(40\%\) come from machine \(B\).
Machine \(A\) produces defective items with probability
\[ 0.02, \]while machine \(B\) produces defective items with probability
\[ 0.05. \]Let \(D\) be the event “defective.”
Then
\[ P(D) = 0.60(0.02)+0.40(0.05) \] \[ = 0.012+0.020 = \boxed{0.032}. \]Conditional Probability Can Update Information
Conditioning is a mathematical representation of learning information.
Before learning \(B\), we may assess
\[ P(A). \]After learning \(B\), the relevant probability becomes
\[ P(A\mid B). \]These values can be smaller, larger, or equal depending on the relationship between the events.
Positive Association
If learning \(B\) makes \(A\) more likely, then
\[ P(A\mid B)>P(A). \]This indicates a positive association between the events in the probability model.
Negative Association
If learning \(B\) makes \(A\) less likely, then
\[ P(A\mid B)No Change Under Conditioning
If
\[ P(A\mid B)=P(A), \]then learning \(B\) does not change the probability of \(A\).
This relationship leads directly to the concept of independence.
Independence Preview
If \(A\) and \(B\) are independent and \(P(B)>0\), then
\[ \boxed{ P(A\mid B)=P(A) }. \]Likewise, when \(P(A)>0\),
\[ P(B\mid A)=P(B). \]The next lesson develops independence systematically.
Example: Independent Coin Tosses
A fair coin is tossed twice.
Let \(A\) be “the second toss is heads” and \(B\) be “the first toss is heads.”
Then
\[ P(A)=\frac12. \]Given that the first toss is heads, the second toss is still equally likely to be heads or tails:
\[ P(A\mid B)=\frac12. \]Conditioning on \(B\) does not change the probability of \(A\).
Example: Dependent Card Draws
Two cards are drawn without replacement.
Let \(A\) be “the second card is an ace” and \(B\) be “the first card is an ace.”
Before any card is drawn,
\[ P(A)=\frac4{52}. \]Given that the first card is an ace,
\[ P(A\mid B)=\frac3{51}. \]The probability changes, so the events are dependent.
Conditional Probability and Geometric Models
Conditioning can also restrict a continuous geometric sample space.
In a uniform area model,
\[ P(A\mid B) = \frac{\operatorname{Area}(A\cap B)} {\operatorname{Area}(B)} \]whenever \(B\) has positive area.
Example: Conditional Area
A point is selected uniformly from the unit square.
Let
\[ A=\{x\le0.4\} \]and
\[ B=\{y\le0.5\}. \]The area of \(B\) is
\[ 0.5. \]The intersection \(A\cap B\) is a
\[ 0.4\times0.5 \]rectangle with area
\[ 0.2. \]Therefore,
\[ P(A\mid B) = \frac{0.2}{0.5} = \boxed{0.4}. \]Conditioning Can Be Nested
In a multistage experiment, later probabilities can condition on several earlier events.
For example,
\[ P(C\mid A\cap B) \]means the probability of \(C\) given that both \(A\) and \(B\) have already occurred.
This structure appears in the chain rule:
\[ P(A\cap B\cap C) = P(A)P(B\mid A)P(C\mid A\cap B). \]Example: Three Draws Without Replacement
Three cards are drawn without replacement.
The probability that all three are aces is
\[ P(A_1) P(A_2\mid A_1) P(A_3\mid A_1\cap A_2). \]Thus,
\[ \boxed{ \frac4{52}\cdot\frac3{51}\cdot\frac2{50} }. \]Each conditional probability reflects the updated deck after the previous information is known.
Conditional Probability and Diagnostic Reasoning
Many real-world applications involve two different conditional probabilities that are easy to confuse.
For example,
\[ P(\text{positive test}\mid\text{condition}) \]is generally not the same as
\[ P(\text{condition}\mid\text{positive test}). \]The first conditions on the condition. The second conditions on the test result.
Base Rates Matter
Reversing a conditional probability requires information about how common the conditioning events are.
A high value of
\[ P(B\mid A) \]does not by itself imply a high value of
\[ P(A\mid B). \]The overall probabilities of \(A\) and \(B\) also matter.
Bayes' Relationship Preview
Since
\[ P(A\cap B) = P(A)P(B\mid A) \]and
\[ P(A\cap B) = P(B)P(A\mid B), \]we can equate the two expressions:
\[ P(A)P(B\mid A) = P(B)P(A\mid B). \]Solving for \(P(A\mid B)\) gives
\[ \boxed{ P(A\mid B) = \frac{P(B\mid A)P(A)}{P(B)} }. \]This relationship is commonly known as Bayes' rule.
Example: Reverse Conditioning
Suppose
\[ P(A)=0.10, \qquad P(B\mid A)=0.80, \qquad P(B)=0.20. \]Then
\[ P(A\mid B) = \frac{0.80(0.10)}{0.20} = \boxed{0.40}. \]The reverse conditional is not \(0.80\); the base probabilities matter.
Common Error: Reversing the Conditional
In general,
\[ \boxed{ P(A\mid B)\neq P(B\mid A) }. \]Always identify which event appears after the conditioning bar.
Common Error: Using the Full Sample Space as the Denominator
Once \(B\) is known to have occurred, outcomes outside \(B\) should not remain in the denominator.
In an equally likely finite model,
\[ \boxed{ P(A\mid B) = \frac{|A\cap B|}{|B|} }, \]not
\[ \frac{|A\cap B|}{|S|}. \]Common Error: Using \(P(A)/P(B)\)
Conditional probability is not generally
\[ \frac{P(A)}{P(B)}. \]The numerator must be the overlap:
\[ \boxed{ P(A\mid B) = \frac{P(A\cap B)}{P(B)} }. \]Common Error: Ignoring Changed Conditions
In without-replacement problems, the first result changes the later sample space.
Conditional probabilities must use the updated numbers of favorable and total outcomes.
Common Error: Assuming Conditioning Always Changes Probability
Sometimes
\[ P(A\mid B)=P(A). \]In that case, the information \(B\) does not change the probability of \(A\).
This is precisely the pattern associated with independence.
Common Error: Treating Association as Causation
If
\[ P(A\mid B)\neq P(A), \]then \(A\) and \(B\) are associated in the probability model.
This alone does not establish that \(B\) causes \(A\), or that \(A\) causes \(B\).
Probability describes relationships among events; causal conclusions require additional assumptions and evidence.
Common Error: Ignoring Base Rates
A conditional rate such as
\[ P(B\mid A) \]cannot usually be reversed without knowing the prevalence of \(A\) and the overall probability of \(B\).
Base rates are part of the denominator structure in reverse-conditioning problems.
A Reliable Procedure for Conditional Probability
- Identify the target event.
- Identify the given or conditioning event.
- Place the conditioning event after the vertical bar.
- Restrict attention to outcomes satisfying the conditioning event.
- Find the overlap between the target and conditioning events.
- Use \(P(A\mid B)=P(A\cap B)/P(B)\).
- For equally likely finite outcomes, use \(|A\cap B|/|B|\).
- For tables, use the conditioning row or column total as the denominator.
- For sequential problems, update the sample space after each known result.
- Check whether the conditional probability is between \(0\) and \(1\).
Example: Full Table Workflow
Suppose \(200\) customers are classified as follows:
| Subscribed | Not Subscribed | Total | |
|---|---|---|---|
| New Customers | 45 | 75 | 120 |
| Returning Customers | 50 | 30 | 80 |
| Total | 95 | 105 | 200 |
Given that a customer is returning, the effective sample space contains \(80\) customers.
Of those, \(50\) are subscribed.
Therefore,
\[ \boxed{ P(\text{subscribed}\mid\text{returning}) = \frac{50}{80} = \frac58 }. \]Practice 1: Basic Conditional Formula
Suppose
\[ P(A\cap B)=0.24 \]and
\[ P(B)=0.60. \]Find \(P(A\mid B)\).
Practice 2: Fair Die
A fair six-sided die is rolled.
Given that the result is even, find the probability that it is greater than \(4\).
Practice 3: Uniform Integer Selection
One integer is selected uniformly from
\[ \{1,2,\ldots,24\}. \]Given that it is divisible by \(3\), find the probability that it is also divisible by \(6\).
Practice 4: Reverse the Condition
Suppose
\[ P(A)=0.40, \qquad P(B)=0.50, \qquad P(A\cap B)=0.20. \]Find both \(P(A\mid B)\) and \(P(B\mid A)\).
Practice 5: Multiplication Rule
Suppose
\[ P(A)=0.70 \]and
\[ P(B\mid A)=0.30. \]Find \(P(A\cap B)\).
Practice 6: Cards Without Replacement
One card is drawn from a standard deck and is known to be a face card.
Find the probability that it is a king.
Practice 7: Second Draw Given the First
Two cards are drawn without replacement.
Given that the first card is red, find the probability that the second card is also red.
Practice 8: Two-Way Table
A survey gives the following counts:
| Prefers A | Prefers B | Total | |
|---|---|---|---|
| Group 1 | 24 | 16 | 40 |
| Group 2 | 18 | 42 | 60 |
| Total | 42 | 58 | 100 |
Find
\[ P(\text{Prefers A}\mid\text{Group 1}). \]Practice 9: Reverse Table Conditional
Using the table in Practice 8, find
\[ P(\text{Group 1}\mid\text{Prefers A}). \]Practice 10: Total Probability
Suppose
\[ P(A)=0.65, \] \[ P(B\mid A)=0.10, \]and
\[ P(B\mid A^c)=0.30. \]Find \(P(B)\).
Practice 11: Geometric Conditioning
A point is selected uniformly from the unit square.
Let
\[ A=\{x\le0.3\} \]and
\[ B=\{y\le0.6\}. \]Find \(P(A\mid B)\).
Practice 12: Identify the Error
A student is told that
\[ P(\text{positive test}\mid\text{condition})=0.95 \]and concludes
\[ P(\text{condition}\mid\text{positive test})=0.95. \]Explain the error.
Practice Answers
Practice 1
\[ P(A\mid B) = \frac{0.24}{0.60} = \boxed{0.40}. \]Practice 2
The even outcomes are
\[ \{2,4,6\}. \]Only \(6\) is greater than \(4\), so
\[ \boxed{ P(\text{greater than }4\mid\text{even}) = \frac13 }. \]Practice 3
The multiples of \(3\) are
\[ 3,6,9,12,15,18,21,24, \]giving \(8\) conditioned outcomes.
The multiples of \(6\) among them are
\[ 6,12,18,24, \]so
\[ \boxed{ P(\text{divisible by }6\mid\text{divisible by }3) = \frac48 = \frac12 }. \]Practice 4
\[ P(A\mid B) = \frac{0.20}{0.50} = \boxed{0.40}, \]while
\[ P(B\mid A) = \frac{0.20}{0.40} = \boxed{0.50}. \]Practice 5
\[ P(A\cap B) = P(A)P(B\mid A) = 0.70(0.30) = \boxed{0.21}. \]Practice 6
A standard deck has \(12\) face cards and \(4\) kings.
Therefore,
\[ \boxed{ P(\text{king}\mid\text{face card}) = \frac4{12} = \frac13 }. \]Practice 7
After one red card is drawn, \(25\) red cards remain among \(51\) cards:
\[ \boxed{ P(\text{second red}\mid\text{first red}) = \frac{25}{51} }. \]Practice 8
Group 1 contains \(40\) people, of whom \(24\) prefer A:
\[ \boxed{ P(\text{Prefers A}\mid\text{Group 1}) = \frac{24}{40} = 0.60 }. \]Practice 9
There are \(42\) people who prefer A, of whom \(24\) are in Group 1:
\[ \boxed{ P(\text{Group 1}\mid\text{Prefers A}) = \frac{24}{42} = \frac47 }. \]Practice 10
First,
\[ P(A^c)=0.35. \]Then
\[ P(B) = 0.65(0.10)+0.35(0.30) \] \[ = 0.065+0.105 = \boxed{0.17}. \]Practice 11
The area of \(B\) is
\[ 0.6. \]The intersection has area
\[ 0.3(0.6)=0.18. \]Therefore,
\[ P(A\mid B) = \frac{0.18}{0.6} = \boxed{0.30}. \]Practice 12
The two conditional probabilities reverse the roles of the events:
\[ P(\text{positive}\mid\text{condition}) \]conditions on having the condition, while
\[ P(\text{condition}\mid\text{positive}) \]conditions on receiving a positive result.
They are generally different. The reverse probability also depends on the prevalence of the condition and the overall frequency of positive tests.
What to Remember
The definition of conditional probability is
\[ \boxed{ P(A\mid B) = \frac{P(A\cap B)}{P(B)} }, \qquad P(B)>0. \]In an equally likely finite sample space,
\[ \boxed{ P(A\mid B) = \frac{|A\cap B|}{|B|} }. \]Rearranging gives the multiplication rule
\[ \boxed{ P(A\cap B) = P(B)P(A\mid B) = P(A)P(B\mid A) }. \]- Conditioning restricts the effective sample space.
- The event after the vertical bar determines the denominator.
- \(P(A\mid B)\) and \(P(B\mid A)\) are generally different.
- Two-way tables use the conditioning row or column total as the denominator.
- Tree diagrams multiply along conditional paths.
- Without-replacement probabilities change because the sample space changes.
- If conditioning does not change probability, independence is suggested.
The most important habit is: once information is given, rebuild the probability calculation inside the outcomes consistent with that information.
Where We Go Next
Conditional probability measures how information about one event changes the probability of another. Sometimes that information produces no change at all.
The next lesson, Independence, develops the precise conditions under which events do not alter one another's probabilities and connects independence to multiplication rules, repeated trials, and probabilistic modeling.