Probability · Lesson 1 of 5

Conditional Probability: Updating the Sample Space

“Use new information to identify the possible outcomes and calculate a conditional probability.”

Learning Objectives

• Interpret a condition as a restriction on the sample space. • Derive and apply conditional probability with a positive-probability condition. • Use conditional addition and complement rules with clear event notation. • Distinguish equally likely outcomes from outcomes carrying different weights. • Describe random variables as numerical functions on a sample space.

What changes when information arrives?

Imagine tossing three fair coins and asking whether at least two show heads. Now suppose someone tells you that the first coin shows a tail. The coins have already landed, but the information changes which outcomes you should consider. Conditional probability measures uncertainty after such information is taken into account.

A sample space $S$ contains all possible outcomes of an experiment. An event is a subset of those outcomes. We write $E \cap F$ for outcomes in both events, $E \cup F$ for outcomes in at least one, and $E^{'}$ for outcomes outside $E$ . When elementary outcomes are equally likely, event probability is the number of favourable outcomes divided by the total number. Otherwise we add their individual probability weights.

For three fair independent coin tosses, $S = {H H H, H H T, H T H, H T T, T H H, T H T, T T H, T T T}$ . Let $E$ mean at least two heads and $F$ mean a tail on the first toss. Four outcomes satisfy $E$ , so $P (E) = 1/2$ . Four satisfy $F$ , but only $T H H$ satisfies both. Once $F$ is known, the four outcomes outside $F$ are removed from consideration.

Restricting the possible outcomes— Look within the lower row, rather than continuing to divide by all eight outcomes.

The remaining possibilities are $T H H, T H T, T T H, T T T$ . They are equally likely, and only one has at least two heads. Thus the new probability is $1/4$ . The notation $P (E ∣ F)$ is read as the probability of $E$ given $F$ . The event after the vertical bar tells us which information is already known.

From restricted counting to a general formula

The successful outcomes must satisfy both the requested event and the condition. The denominator must count all outcomes consistent with the condition. This distinction is the heart of the formula; changing the denominator is not an arbitrary algebraic trick.

For equally likely outcomes, the restricted fraction is $n (E \cap F) / n (F)$ . Dividing its numerator and denominator by $n (S)$ converts this to $P (E \cap F) / P (F)$ . This second expression also works when elementary outcomes have unequal probabilities, because it compares probability weights rather than numbers of outcomes.

Definition

Conditional probability

If $E$ and $F$ belong to the same experiment and $P (F) > 0$ , then $P (E ∣ F)$ is the probability of $E$ after conditioning on $F$ . It is the proportion of the probability weight of $F$ that also lies in $E$ .

Conditional probabilityLaTeX

P (E ∣ F) = \frac{P ( E \cap F )}{P ( F )}, P (F) > 0

The numerator is the joint probability of

E

and

F

. The denominator is the probability of the known condition

F

. Since

E \cap F

is contained in

F

, the ratio lies between

0

and

1

The condition must have positive probability

If $P (F) = 0$ , this definition would require division by zero and does not define $P (E ∣ F)$ . In the finite equally likely experiments here, any nonempty event has positive probability. Do not assume that nonemptiness alone guarantees positive probability in every possible probability model.

Example — Calculate from joint probabilities

Problem
Suppose $P (A) = 7/13$ , $P (B) = 9/13$ , and $P (A \cap B) = 4/13$ . Find $P (A ∣ B)$ .

1.The known information is $B$ , so its probability belongs in the denominator.
2.Use $P (A ∣ B) = (4/13) / (9/13) = 4/9$ .
3.This ratio describes the part of the probability weight of $B$ that lies in $A$ . The value $P (A) = 7/13$ is not the denominator.

Recognising the condition in a story

Words such as given, known, observed, and among often signal conditioning. First name the requested event, then name the information that is known. In family and dice problems, preserve the order or identity of outcomes so that the sample points really are equally likely.

Example — Two children and at least one boy

Problem
Model two children by equally likely ordered outcomes $bb, b g, g b, g g$ . Given that at least one is a boy, find the probability that both are boys.

1.Let $E = {bb}$ and $F = {bb, b g, g b}$ . The condition excludes only $g g$ .
2.The three remaining ordered outcomes are equally likely. One satisfies $E$ . Equivalently, $P (E \cap F) = 1/4$ and $P (F) = 3/4$ .
3.Therefore $P (E ∣ F) = (1/4) / (3/4) = 1/3$ . If the information instead specifies that the younger child is a boy, the remaining set is different and the answer is $1/2$ .

Example — A numbered card

Problem
A card is selected uniformly from cards numbered $1$ to $10$ . Given that its number exceeds $3$ , find the probability that it is even.

1.The condition leaves the seven numbers $4, 5, 6, 7, 8, 9, 10$ .
2.The favourable numbers within that set are $4, 6, 8, 10$ , giving four outcomes.
3.Thus the conditional probability is $4/7$ . Using all ten cards in the denominator would ignore the information.

Example — A percentage within a group

Problem
A school has $1000$ students, including $430$ girls. Ten percent of the girls are in Class XII. Given that a randomly selected student is a girl, what is the probability that the student is in Class XII?

1.There are $0.10 \times 430 = 43$ girls in Class XII. Let $E$ mean Class XII and $F$ mean girl.
2.Then $P (E \cap F) = 43/1000$ and $P (F) = 430/1000$ .
3.The ratio is $43/430 = 0.10$ . The given ten percent already describes the conditional group; it is not ten percent of all students.

Example — Three throws with the first two known

Problem
A fair die is thrown three times. Given that the first two results are $6$ and $5$ , find the probability that the third is $4$ .

1.The complete experiment has $6^{3} = 216$ equally likely ordered outcomes.
2.The condition leaves $(6, 5, 1), (6, 5, 2), \dots, (6, 5, 6)$ . Only $(6, 5, 4)$ is favourable.
3.The required probability is $(1/216) / (6/216) = 1/6$ . Knowing the earlier throws does not change this third-throw probability.

Example — An observed sum of six

Problem
Two fair dice are thrown. Given that their sum is $6$ , find the probability that a $4$ appears at least once.

1.The condition leaves $(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)$ , five equally likely ordered outcomes.
2.The requested event holds for $(2, 4)$ and $(4, 2)$ . The outcomes $(4, 1)$ and $(4, 6)$ are irrelevant because they do not satisfy the condition.
3.Hence the conditional probability is $2/5$ . Writing the restricted sample space is often the quickest way to avoid errors.

Conditional probabilities obey familiar rules

Once $F$ is fixed, conditional probability behaves like an ordinary probability measure on the remaining possibilities. The whole remaining space has probability one. Addition and complement rules still work, provided the same condition is used throughout.

Since $S \cap F = F$ and $F \cap F = F$ , dividing by $P (F)$ gives $P (S ∣ F) = P (F ∣ F) = 1$ . For an addition rule, distribute intersection: $(A \cup B) \cap F = (A \cap F) \cup (B \cap F)$ . Apply the ordinary addition rule to this union and divide every term by $P (F)$ .

Conditional additionLaTeX

P ((A \cup B) ∣ F) = P (A ∣ F) + P (B ∣ F) - P ((A \cap B) ∣ F)

Subtract the overlap because it appears in both individual terms. If

A

and

B

are disjoint, the overlap term is zero.

The events $E$ and $E^{'}$ are disjoint and cover $S$ . Their conditional probabilities therefore add to $P (S ∣ F) = 1$ . Rearranging gives the conditional complement rule.

Conditional complementLaTeX

P (E^{'} ∣ F) = 1 - P (E ∣ F)

The complement is taken in the original sample space; within

F

, it is the portion of

F

that does not satisfy

E

Reversing the bar changes the question

Generally $P (A ∣ B)$ and $P (B ∣ A)$ have different denominators. Even if they happen to be equal, this does not always imply $P (A) = P (B)$ . If $A$ and $B$ are disjoint and both have positive probability, both conditional probabilities are zero regardless of the two marginal probabilities. With positive overlap, equality does imply equal marginals by cancelling $P (A \cap B)$ .

A tree can contain unequally likely leaves

A list of possible outcomes does not establish that they have equal probabilities. If the experiment changes its next action according to an earlier result, different leaves may require different sequences of chances. A probability tree records those sequences clearly.

Example — Toss again or roll a die

Problem
Toss a fair coin. After a head, toss it again; after a tail, roll a fair die. Given that at least one tail occurred, find the probability that the die showed a number greater than $4$ .

1.The leaves are $H H, H T, T 1, T 2, T 3, T 4, T 5, T 6$ . The first two each have probability $(1/2) (1/2) = 1/4$ . Each die leaf has probability $(1/2) (1/6) = 1/12$ .
2.Let $F$ mean at least one tail. It consists of $H T$ and the six die leaves, so $P (F) = 1/4 + 6/12 = 3/4$ . Let $E = {T 5, T 6}$ , so $P (E \cap F) = 2/12 = 1/6$ .
3.Divide the probability weights: $P (E ∣ F) = (1/6) / (3/4) = 2/9$ . Counting two favourable leaves among seven remaining leaves would incorrectly treat those leaves as equally likely.

The coin and die probability tree— Multiply branch probabilities to obtain leaf weights; then add the weights belonging to the condition.

Assigning numbers to outcomes

Events classify outcomes as belonging or not belonging to a set. We can also record a number for every outcome. This gives another precise description of an experiment and explains the brief random-variable idea used in this chapter.

Definition

Random variable

A random variable is a real-valued function defined on the sample space of a random experiment. Each outcome is assigned exactly one real number, although different outcomes may have the same assigned number.

Outcome	Number of heads, X	Heads minus tails, Y
HH	2	2
HT	1	0
TH	1	0
TT	0	−2

For two tosses, $X$ counts heads and $Y$ counts heads minus tails. These are different functions on the same sample space, with $Y = 2 X - 2$ . The value $X = 1$ corresponds to two outcomes, $H T$ and $T H$ . A random variable is a rule applied to an outcome, not an extra coin toss or an additional outcome.

Check your understanding

Focus on the known information before calculating. Each question checks whether you have identified the correct remaining possibilities and the correct probability weights.

Quick check

If $P (E \cap F) = 0.12$ and $P (F) = 0.30$ , what is $P (E ∣ F)$ ?

Quick check

Which condition is required by the ratio definition of $P (E ∣ F)$ ?

Quick check

Given that the younger of two children is a girl, what is the probability that both are girls under the equally likely ordered-outcome model?

Quick check

For $P (E ∣ F) = 0.7$ , what is $P (E^{'} ∣ F)$ ?

Quick check

Why is simply counting leaves wrong in the coin-then-coin-or-die experiment?

Quick check

A random variable $X$ counts heads in two coin tosses. Which event corresponds to $X = 1$ ?

Quick check

If $A \subseteq B$ and $P (B) > 0$ , which statement must hold?

Practice Problems

Given $P (E) = 0.6$ , $P (F) = 0.3$ , and $P (E \cap F) = 0.2$ , find both reverse conditional probabilities.
Find $P (A ∣ B)$ when $P (B) = 0.5$ and $P (A \cap B) = 0.32$ .
Let $P (A) = 0.8$ , $P (B) = 0.5$ , and $P (B ∣ A) = 0.4$ . Find the intersection probability, $P (A ∣ B)$ , and the union probability.
Suppose $2 P (A) = P (B) = 5/13$ and $P (A ∣ B) = 2/5$ . Find $P (A \cup B)$ .
For $P (A) = 6/11$ , $P (B) = 5/11$ , and $P (A \cup B) = 7/11$ , find the intersection and both reverse conditional probabilities.
Three fair coins are tossed. Find (a) probability of a head on the third toss given heads on the first two; (b) at least two heads given at most two heads; (c) at most two tails given at least one tail.
Two fair coins are tossed. Find (a) probability of at least one tail given at least one head; (b) probability of no tails given no heads.
A die is thrown three times. Find the probability of $4$ on the third throw given $6$ then $5$ on the first two.
A mother, father, and son stand in a uniformly random order. Given that the father is in the middle, find the probability that the son is at an end.
A black and a red die are rolled. Find (a) probability of a sum greater than $9$ given a black $5$ ; (b) probability of a sum of $8$ given that the red result is less than $4$ .
For a fair die let $E = {1, 3, 5}$ , $F = {2, 3}$ , $G = {2, 3, 4, 5}$ . Find $P (E ∣ F)$ , $P (F ∣ E)$ , $P (E ∣ G)$ , $P (G ∣ E)$ , $P ((E \cup F) ∣ G)$ , and $P ((E \cap F) ∣ G)$ .
For two children, find the probability that both are girls given (a) the younger is a girl; (b) at least one is a girl. State your model.
A question bank has $300$ easy true/false, $200$ difficult true/false, $500$ easy multiple-choice, and $400$ difficult multiple-choice questions. Given a multiple-choice selection, find the probability that it is easy.
Two dice show different numbers. Find the probability that their sum is $4$ .
Roll a die. If the result is a multiple of $3$ , roll again; otherwise toss a coin. Given that a die showed $3$ at least once, find the probability that a coin showed a tail.
Explain why $P (A ∣ B)$ is undefined by the ratio formula when $P (B) = 0$ , even if $P (A) = 1/2$ .
Find disjoint positive-probability events with unequal probabilities but $P (A ∣ B) = P (B ∣ A)$ . Explain why positive overlap changes the conclusion.
For $A \subseteq B$ with $P (B) > 0$ , prove $P (A ∣ B) \geq P (A)$ and identify when equality can occur.
For two coin tosses, list the values of $X =$ number of heads and $Y =$ heads minus tails. Which outcomes produce $Y = 0$ ?

$P (E ∣ F) = 0.2/0.3 = 2/3$ and $P (F ∣ E) = 0.2/0.6 = 1/3$ .

Key Takeaways

• Conditioning retains the outcomes consistent with the known event and renormalises their probability weights. • Use $P (E ∣ F) = P (E \cap F) / P (F)$ only when $P (F) > 0$ . • Conditional addition and complement rules use the same fixed condition throughout. • Outcome counts work only for equally likely outcomes; probability trees can expose unequal weights. • A random variable assigns a real number to each outcome, and different functions can describe the same experiment.

Start of chapter

Next · Lesson 2

Joint Events, Sequential Trials, and Independence

What changes when information arrives?

From restricted counting to a general formula

Recognising the condition in a story

Conditional probabilities obey familiar rules

A tree can contain unequally likely leaves

Assigning numbers to outcomes

Check your understanding

Quiz

Practice Problems

Key Takeaways