The Central Idea

Suppose \(A\) and \(B\) are events. If we learn that \(B\) has occurred, outcomes outside \(B\) are no longer relevant.

The probability of \(A\) under the condition that \(B\) occurred is written

\[ P(A\mid B). \]

The symbol \(\mid\) is read as given. Thus, \(P(A\mid B)\) means the probability of \(A\) given \(B\).

Conditioning changes the reference set. The new question is not “How much of the whole sample space lies in \(A\)?” but rather “How much of \(B\) also lies in \(A\)?”

The Conditional Probability Formula

If \(P(B)>0\), then

\[ \boxed{P(A\mid B)=\frac{P(A\cap B)}{P(B)}}. \]

The numerator \(P(A\cap B)\) measures the probability of outcomes that satisfy both \(A\) and \(B\).

The denominator \(P(B)\) measures the probability of the event that has become our new reference set.

Why the Formula Works

Once \(B\) is known to have occurred, the event \(B\) becomes the effective sample space. Within that reduced sample space, the favorable outcomes for \(A\) are exactly those in

\[ A\cap B. \]

Dividing by \(P(B)\) rescales the probability of \(B\) to \(1\):

\[ \frac{P(B)}{P(B)}=1. \]

The portion of this rescaled space that also belongs to \(A\) is therefore

\[ \frac{P(A\cap B)}{P(B)}. \]

Conditioning Restricts the Sample Space

Let

\[ S=\{1,2,3,4,5,6\} \]

represent the outcomes of a fair six-sided die.

Define

\[ A=\{2,4,6\} \]

as the event that the result is even, and

\[ B=\{4,5,6\} \]

as the event that the result is greater than \(3\).

If we know \(B\) occurred, the effective sample space is no longer all six outcomes. It is

\[ B=\{4,5,6\}. \]

Of those three outcomes, two are even:

\[ A\cap B=\{4,6\}. \]

Therefore,

\[ \boxed{P(A\mid B)=\frac{2}{3}}. \]

The Same Example Using the Formula

For the die example,

\[ P(A\cap B)=\frac{2}{6} \]

and

\[ P(B)=\frac{3}{6}. \]

Thus,

\[ P(A\mid B) = \frac{P(A\cap B)}{P(B)} = \frac{2/6}{3/6} = \boxed{\frac{2}{3}}. \]

Direct counting in the restricted sample space and the formal probability formula give the same result.

The Order of the Events Matters

In general,

\[ P(A\mid B)\neq P(B\mid A). \]

These expressions use different reference sets.

In the die example,

\[ P(A\mid B)=\frac{2}{3}. \]

But if we instead know that the result is even, then the effective sample space is

\[ A=\{2,4,6\}. \]

Of these three outcomes, two are greater than \(3\):

\[ \{4,6\}. \]

Therefore,

\[ P(B\mid A)=\frac{2}{3}. \]

These happen to be equal in this example, but that equality is accidental, not a general rule.

An Example Where Reversing the Condition Changes the Probability

Suppose a class contains \(30\) students. Of these:

Let \(M\) be the event that a randomly selected student studies mathematics, and \(C\) the event that the student studies computer science.

Among the \(12\) computer science students, \(9\) also study mathematics:

\[ P(M\mid C)=\frac{9}{12}=\boxed{\frac{3}{4}}. \]

Among the \(18\) mathematics students, only \(9\) also study computer science:

\[ P(C\mid M)=\frac{9}{18}=\boxed{\frac{1}{2}}. \]

Therefore,

\[ P(M\mid C)\neq P(C\mid M). \]

The Condition Must Have Positive Probability

The formula

\[ P(A\mid B)=\frac{P(A\cap B)}{P(B)} \]

requires

\[ P(B)>0. \]

If \(P(B)=0\), the denominator is zero, so the elementary ratio definition is not defined.

Later probability theory can treat conditioning in more advanced settings, but this lesson uses the standard elementary definition.

Conditional Probability with Equally Likely Outcomes

When outcomes are equally likely and \(B\) contains at least one outcome, conditional probability can be computed by counting:

\[ \boxed{ P(A\mid B) = \frac{|A\cap B|}{|B|} }. \]

Here \(|B|\) means the number of outcomes in \(B\).

This is the ordinary favorable-outcomes-over-possible-outcomes rule applied to the reduced sample space \(B\).

Example: Drawing a Card

A standard deck contains \(52\) cards. Suppose we know that the selected card is a face card.

There are \(12\) face cards:

\[ 4\text{ jacks}+4\text{ queens}+4\text{ kings}. \]

Of these, \(4\) are kings. Therefore,

\[ P(\text{king}\mid\text{face card}) = \frac{4}{12} = \boxed{\frac{1}{3}}. \]

The denominator is \(12\), not \(52\), because the condition has restricted the relevant outcomes to face cards.

Example: A Condition Can Increase Probability

In a standard deck,

\[ P(\text{king})=\frac{4}{52}=\frac{1}{13}. \]

But after learning that the card is a face card,

\[ P(\text{king}\mid\text{face card})=\frac{1}{3}. \]

The condition makes a king much more plausible because kings form a larger share of the restricted set of face cards than of the full deck.

Example: A Condition Can Decrease Probability

Let \(A\) be the event that a card is red. Unconditionally,

\[ P(A)=\frac{26}{52}=\frac{1}{2}. \]

Now suppose we learn that the card is a spade. Every spade is black, so

\[ P(\text{red}\mid\text{spade})=\boxed{0}. \]

New information can raise, lower, or leave unchanged the probability of an event.

Example: A Condition Can Leave Probability Unchanged

Suppose a fair die is rolled. Let

\[ A=\{\text{even}\} \]

and

\[ B=\{\text{result greater than }2\}. \]

Then

\[ B=\{3,4,5,6\} \]

and the even outcomes inside \(B\) are

\[ \{4,6\}. \]

Therefore,

\[ P(A\mid B)=\frac{2}{4}=\frac{1}{2}. \]

Since

\[ P(A)=\frac{3}{6}=\frac{1}{2}, \]

this particular condition does not change the probability of \(A\). The next lesson develops the formal connection between this phenomenon and independence.

Conditional Probability from a Two-Way Table

Conditional probability is often read directly from a table.

Suppose \(200\) customers are classified by whether they purchased a subscription and whether they used a free trial.

Purchased Did Not Purchase Total
Used Trial 72 48 120
No Trial 24 56 80
Total 96 104 200

Conditioning on a Row

Suppose we want

\[ P(\text{Purchased}\mid\text{Used Trial}). \]

The condition “Used Trial” restricts attention to the \(120\) customers in that row.

Of those \(120\), \(72\) purchased:

\[ P(\text{Purchased}\mid\text{Used Trial}) = \frac{72}{120} = \boxed{0.60}. \]

Conditioning on a Column

Now consider

\[ P(\text{Used Trial}\mid\text{Purchased}). \]

The condition “Purchased” restricts attention to the \(96\) customers in that column.

Of those \(96\), \(72\) used a trial:

\[ P(\text{Used Trial}\mid\text{Purchased}) = \frac{72}{96} = \boxed{0.75}. \]

The same intersection count, \(72\), appears in both calculations, but the denominators differ because the conditions differ.

A Reliable Table Strategy

When computing a conditional probability from a table:

  1. Identify the event after the vertical bar.
  2. Restrict attention to the corresponding row, column, or category.
  3. Use that restricted total as the denominator.
  4. Find the count satisfying both events.
  5. Use that intersection count as the numerator.

The event after the bar controls the denominator.

Conditional Probability from Percentages

Counts are not required. Suppose

\[ P(A\cap B)=0.18 \]

and

\[ P(B)=0.30. \]

Then

\[ P(A\mid B) = \frac{0.18}{0.30} = \boxed{0.60}. \]

The numerator and denominator may be probabilities, proportions, percentages expressed as decimals, or counts measured on the same base.

Conditional Probability and Intersections

The formula can be rearranged:

\[ P(A\mid B)=\frac{P(A\cap B)}{P(B)}. \]

Multiplying both sides by \(P(B)\) gives

\[ \boxed{P(A\cap B)=P(B)P(A\mid B)}. \]

This relationship is one form of a multiplication rule. It appears here because it follows directly from the definition, but multiplication rules will be developed systematically in a later lesson.

The Symmetric Multiplication Relationship

By reversing the roles of \(A\) and \(B\),

\[ P(B\mid A)=\frac{P(A\cap B)}{P(A)}. \]

Therefore,

\[ P(A\cap B)=P(A)P(B\mid A). \]

Combining the two forms gives

\[ \boxed{ P(B)P(A\mid B) = P(A\cap B) = P(A)P(B\mid A) }. \]

This equality helps explain why reversing a condition usually changes the conditional probability: the denominators \(P(A)\) and \(P(B)\) need not be equal.

Conditional Complements

Once \(B\) is fixed as the condition, \(A\) and \(A^c\) still form complementary possibilities inside \(B\).

Therefore,

\[ \boxed{P(A^c\mid B)=1-P(A\mid B)}. \]

For example, if

\[ P(A\mid B)=0.72, \]

then

\[ P(A^c\mid B)=1-0.72=\boxed{0.28}. \]

Conditional Probabilities Still Obey Probability Rules

For a fixed condition \(B\) with \(P(B)>0\), conditional probabilities act like ordinary probabilities within the restricted sample space.

In particular,

\[ 0\leq P(A\mid B)\leq 1, \] \[ P(B\mid B)=1, \]

and

\[ P(\varnothing\mid B)=0. \]

The probability system has not been abandoned; it has been renormalized within \(B\).

Conditional Addition Within a Fixed Condition

For events \(A\), \(C\), and a condition \(B\) with \(P(B)>0\),

\[ P(A\cup C\mid B) = P(A\mid B)+P(C\mid B)-P(A\cap C\mid B). \]

If \(A\) and \(C\) are mutually exclusive within the conditioned setting, then

\[ P(A\cup C\mid B) = P(A\mid B)+P(C\mid B). \]

This is the familiar addition rule applied after restriction to \(B\).

Example: Manufacturing Inspection

A plant produces components on two shifts. Suppose \(40\%\) of all components come from the night shift, and \(3\%\) of all components are both from the night shift and defective.

Let \(N\) mean night shift and \(D\) mean defective. Then

\[ P(N)=0.40 \]

and

\[ P(D\cap N)=0.03. \]

The probability that a component is defective given that it came from the night shift is

\[ P(D\mid N) = \frac{0.03}{0.40} = \boxed{0.075}. \]

Thus \(7.5\%\) of night-shift production is defective.

Joint Percentage Versus Conditional Percentage

In the manufacturing example,

\[ P(D\cap N)=0.03 \]

means \(3\%\) of all components are both defective and produced on the night shift.

By contrast,

\[ P(D\mid N)=0.075 \]

means \(7.5\%\) of the night-shift components are defective.

Joint probability and conditional probability answer different questions.

Conditional Probability from a Tree

A probability tree can represent stages of information. Suppose a company receives orders from two customer groups:

\[ P(G_1)=0.60,\qquad P(G_2)=0.40. \]

Suppose further that the conditional probability of a repeat purchase is

\[ P(R\mid G_1)=0.70 \]

and

\[ P(R\mid G_2)=0.45. \]

The second-stage branch probabilities are conditional probabilities. They describe behavior inside the subgroup selected at the first stage.

Tree multiplication and aggregation will be developed more fully in later lessons on multiplication rules and total probability.

Conditional Probability from Sequential Sampling

A box contains \(5\) red balls and \(3\) blue balls. Two balls are drawn without replacement.

Suppose the first ball is known to be red. Then \(7\) balls remain: \(4\) red and \(3\) blue.

Therefore,

\[ P(\text{second red}\mid\text{first red}) = \boxed{\frac{4}{7}}. \]

The condition changes the composition of the remaining population.

With Replacement Versus Without Replacement

If the first red ball were replaced before the second draw, the box would return to its original composition:

\[ 5\text{ red},\qquad 3\text{ blue}. \]

Then

\[ P(\text{second red}\mid\text{first red and replaced}) = \frac{5}{8}. \]

Without replacement, the conditional probability was

\[ \frac{4}{7}. \]

Conditioning must reflect the actual information and the actual sampling process.

Natural-Language Clues for Conditioning

Conditional probability frequently appears in phrases such as:

Each phrase signals that the denominator may be a subgroup rather than the entire population.

Translate the Language Before Calculating

Consider the statement: “Among customers who renewed, what proportion used automatic payment?”

Let \(A\) mean used automatic payment and \(R\) mean renewed.

The word among makes renewals the reference group, so the mathematical question is

\[ \boxed{P(A\mid R)}. \]

Writing the symbolic form first helps prevent denominator errors.

Conditional Probability and Population Subgroups

In applied work, conditioning often means comparing rates inside subpopulations.

Examples include:

The same mathematical structure appears across all of these settings.

Conditioning Is an Information Update

Before learning \(B\), probability is distributed across the entire sample space.

After learning \(B\), outcomes outside \(B\) are eliminated from consideration and the probabilities inside \(B\) are rescaled so that they sum to \(1\).

This idea is the foundation for later work with Bayes' theorem, likelihoods, diagnostic testing, and sequential updating.

Common Error: Using the Entire Sample Space as the Denominator

Suppose \(9\) of \(12\) computer science students also study mathematics, within a class of \(30\).

The conditional probability is

\[ P(M\mid C)=\frac{9}{12}, \]

not

\[ \frac{9}{30}. \]

Once \(C\) is given, only the \(12\) computer science students belong in the denominator.

Common Error: Reversing the Condition

The expressions

\[ P(A\mid B) \]

and

\[ P(B\mid A) \]

usually differ.

Read the notation in words before calculating:

Common Error: Confusing Intersection with Conditioning

The event

\[ A\cap B \]

means that both events occur.

The conditional probability

\[ P(A\mid B) \]

measures how common \(A\) is within \(B\).

They are connected by

\[ P(A\mid B)=\frac{P(A\cap B)}{P(B)}, \]

but they are not the same quantity.

Common Error: Treating “Given” as Causation

Conditional probability describes a relationship between information and probability. It does not, by itself, establish that one event causes another.

A large value of

\[ P(A\mid B) \]

means that \(A\) is common among cases where \(B\) occurs. Additional reasoning or evidence is required to make a causal claim.

Common Error: Dividing by the Wrong Event

From

\[ P(A\mid B)=\frac{P(A\cap B)}{P(B)}, \]

the denominator is always the probability of the event after the vertical bar.

A useful memory aid is:

\[ \boxed{\text{condition} \longrightarrow \text{denominator}}. \]

Common Error: Ignoring Whether the Condition Is Possible

Before using

\[ P(A\mid B)=\frac{P(A\cap B)}{P(B)}, \]

verify that

\[ P(B)>0. \]

In elementary probability, conditioning on an event of probability zero is outside the scope of this ratio definition.

A Reliable Procedure

  1. Identify the event being requested.
  2. Identify the condition—the event after the vertical bar.
  3. Restrict attention to outcomes satisfying the condition.
  4. Find the outcomes satisfying both the requested event and the condition.
  5. Compute the appropriate ratio.
  6. Check that the result lies between \(0\) and \(1\).
  7. Interpret the result using the conditioned group as the reference population.

Example: Full Workflow from Counts

A survey of \(500\) customers finds:

Find the probability that a customer uses the mobile app given that the customer subscribes to premium service.

Let \(M\) mean mobile app and \(P\) mean premium.

The condition is \(P\), so the denominator is \(160\). The intersection contains \(100\) customers.

\[ P(M\mid P) = \frac{100}{160} = \boxed{0.625}. \]

Thus \(62.5\%\) of premium subscribers use the mobile app.

Example: Full Workflow from Probabilities

Suppose

\[ P(A)=0.55,\qquad P(B)=0.40,\qquad P(A\cap B)=0.22. \]

Then

\[ P(A\mid B) = \frac{0.22}{0.40} = \boxed{0.55}. \]

Also,

\[ P(B\mid A) = \frac{0.22}{0.55} = \boxed{0.40}. \]

This numerical pattern will become important when independence is developed in the next lesson.

Practice 1: Restricted Sample Space

A fair die is rolled. Given that the result is at least \(3\), find the probability that the result is even.

Practice 2: Cards

A card is drawn from a standard \(52\)-card deck. Given that the card is red, find the probability that it is a heart.

Practice 3: Counts

In a group of \(80\) employees, \(50\) completed safety training, \(32\) completed equipment training, and \(24\) completed both. Find

\[ P(\text{safety}\mid\text{equipment}). \]

Practice 4: Reverse the Condition

Using the employee data from Practice 3, find

\[ P(\text{equipment}\mid\text{safety}). \]

Practice 5: Probabilities

Suppose

\[ P(A\cap B)=0.12 \]

and

\[ P(B)=0.30. \]

Find \(P(A\mid B)\).

Practice 6: Conditional Complement

If

\[ P(A\mid B)=0.68, \]

find \(P(A^c\mid B)\).

Practice 7: Without Replacement

A bag contains \(6\) green tokens and \(4\) yellow tokens. Two tokens are drawn without replacement. Given that the first token is green, find the probability that the second token is green.

Practice 8: Interpret a Table

A retailer records the following \(150\) transactions:

Used Coupon No Coupon Total
Online 36 54 90
In Store 18 42 60
Total 54 96 150

Find

\[ P(\text{Used Coupon}\mid\text{Online}). \]

Practice 9: Reverse the Table Condition

Using the table from Practice 8, find

\[ P(\text{Online}\mid\text{Used Coupon}). \]

Practice 10: Identify the Error

A population contains \(200\) people. Of these, \(80\) are in group \(B\), and \(30\) are in both \(A\) and \(B\).

A student writes

\[ P(A\mid B)=\frac{30}{200}=0.15. \]

Explain the error and give the correct probability.

Practice Answers

Practice 1

Given that the result is at least \(3\), the possible outcomes are

\[ \{3,4,5,6\}. \]

The even outcomes are \(4\) and \(6\), so

\[ \boxed{\frac{2}{4}=\frac{1}{2}}. \]

Practice 2

There are \(26\) red cards, of which \(13\) are hearts:

\[ \boxed{P(\text{heart}\mid\text{red})=\frac{13}{26}=\frac{1}{2}}. \]

Practice 3

The condition is equipment training, so the denominator is \(32\):

\[ \boxed{ P(\text{safety}\mid\text{equipment}) = \frac{24}{32} = \frac{3}{4} }. \]

Practice 4

The condition is now safety training, so the denominator is \(50\):

\[ \boxed{ P(\text{equipment}\mid\text{safety}) = \frac{24}{50} = 0.48 }. \]

Practice 5

\[ P(A\mid B) = \frac{0.12}{0.30} = \boxed{0.40}. \]

Practice 6

\[ P(A^c\mid B) = 1-P(A\mid B) = 1-0.68 = \boxed{0.32}. \]

Practice 7

After one green token is removed, \(9\) tokens remain, including \(5\) green tokens:

\[ \boxed{ P(\text{second green}\mid\text{first green}) = \frac{5}{9} }. \]

Practice 8

Among the \(90\) online transactions, \(36\) used a coupon:

\[ \boxed{ P(\text{Used Coupon}\mid\text{Online}) = \frac{36}{90} = 0.40 }. \]

Practice 9

Among the \(54\) transactions that used a coupon, \(36\) were online:

\[ \boxed{ P(\text{Online}\mid\text{Used Coupon}) = \frac{36}{54} = \frac{2}{3} }. \]

Practice 10

The denominator must be the number of people in the conditioned group \(B\), not the entire population.

\[ P(A\mid B) = \frac{30}{80} = \boxed{0.375}. \]

What to Remember

Conditional probability changes the reference set according to known information.

\[ \boxed{ P(A\mid B)=\frac{P(A\cap B)}{P(B)} } \qquad\text{for }P(B)>0. \]

The core habit is: identify the conditioned group before doing any arithmetic.

Where We Go Next

Conditional probability gives a precise way to measure how information changes probability. Sometimes learning that one event occurred changes the probability of another event, and sometimes it does not.

The next lesson, Independence and Dependence, develops the formal criteria for distinguishing these two situations and connects them directly to conditional probability.