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Conditional Probability and Independence

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Conditional Probability and Independence

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14 min read

2026-09-12

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Now · Before you begin

Conditional Probability and Independence

Conditional probability asks: among the cases consistent with what we know, how often does the event we want occur? The difficult part is usually choosing the right group to count, not dividing two numbers.

This chapter moves from counting in a small table to conditional notation, independence, repeated draws, and weighted probabilities. Work through the examples before the checks. If you already know the basics, start at Question-pattern anatomy and return to an explanation when your reasoning breaks.

Practice, not an assessment

All datasets and exercises in this revision are hypothetical teaching examples. They are not current graded questions or extracted paper keys. The week label comes from the imported note; it does not establish the scope of your upcoming exam.

Before you begin

You need fractions as proportions and the idea of an event, a set of possible outcomes. If a fair die is rolled, the sample space is

\Omega=\{1,2,3,4,5,6\}

and the event “even” is

\{2,4,6\}

. The symbol

\Omega

(capital omega) names the set of all possible outcomes here.

An intersection contains outcomes in both events; a union contains outcomes in either event, including both. Review probability basics, addition and multiplication rules, or contingency tables if these ideas feel unfamiliar.

A quick prerequisite check: among 12 equally likely tickets, 3 are blue. Their proportion is

3/12=1/4

. If we restrict attention to 6 tickets, of which 3 are blue, the proportion becomes

3/6=1/2

. The blue count has not changed; the group we compare it with has.

The denominator tells you which world you are in

Imagine 40 study-club members. Each belongs to exactly one session and either brings a laptop or does not. Select one member uniformly at random, so each member has probability

1/40

of being selected.

Session	Laptop	No laptop	Total
Morning	12	8	20
Evening	6	14	20
Total	18	22	40

Let

A

be “the selected member brings a laptop” and

B

be “the selected member attends the morning session.” These letters name events, not counts.

Without further information, 18 of the 40 members bring laptops. Therefore

P(A)=18/40=0.45

. The notation

P(A)

reads “the probability of event A.”

Now someone tells us that the selected member attends the morning session. We can rule out the entire evening row. Among the 20 remaining members, 12 bring laptops:

P(A\mid B)=\frac{12}{20}=0.6.

Read

\mid

as given: “the probability of A given B.” It is a condition, not division and not an instruction to multiply. The word conditional is useful literally: the probability is calculated under the stated condition.

Nothing about the members changed. Our information changed which outcomes remain eligible.

Joint, conditional, and reversed conditional

The same table answers three different questions:

Both laptop and morning: $P(A\cap B)=12/40$ . We select from everyone and require both properties. The intersection symbol $\cap$ reads “and” here.
Laptop given morning: $P(A\mid B)=12/20$ . We select from the morning row.
Morning given laptop: $P(B\mid A)=12/18$ . We select from the laptop column.

The favourable count is 12 each time, but the denominator changes. Although

A\cap B=B\cap A

, the two conditional probabilities need not be equal.

Read the condition first

P(A\mid B)

, begin with the event to the right of the bar. Say “among B, how many are also A?” Then find the denominator. This works even when the question's sentence mentions A first.

From counts to the definition

For equally likely outcomes, write

n(B)

for the number of outcomes in

B

. Restricting to

B

gives the proportion

n(A\cap B)/n(B)

If there are

N

equally likely outcomes overall, then

P(A\cap B)=n(A\cap B)/N

and

P(B)=n(B)/N

. Dividing these cancels the common

N

. This motivates the general definition:

P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)>0.

This definition also applies when outcomes have unequal probabilities. In that case use their probability weights, not unweighted counts.

The restriction

P(B)>0

matters because division by zero is undefined. Also,

A\cap B

is contained in

B

, so

0\leq P(A\cap B)\leq P(B)

. Consequently the ratio is always between 0 and 1. An answer larger than 1 signals a wrong denominator, inconsistent data, or arithmetic error.

What if the condition has probability zero?

The elementary ratio above does not define conditioning on

B

when

P(B)=0

. In a finite model where every outcome has positive probability, a zero-probability event is empty. In continuous models, an individual value can have probability zero without being logically impossible. More advanced definitions handle some such conditioning problems; substituting zero into this ratio does not.

When the joint probability is not given

Suppose a hypothetical model gives

P(A)=0.4

P(B)=0.5

, and

P(A\cup B)=0.7

. The union symbol

\cup

means “A or B, including both.” Use the addition rule to recover the overlap:

P(A\cap B)=P(A)+P(B)-P(A\cup B)=0.4+0.5-0.7=0.2.

Then condition on

B

P(A\mid B)=\frac{0.2}{0.5}=0.4.

The subtraction is necessary because adding

P(A)

and

P(B)

counts their overlap twice. Do not put the union probability in the numerator of a conditional probability.

Independence means the information does not change the probability

Two events are independent when their joint probability factors as

P(A\cap B)=P(A)P(B).

This is a definition, not a shortcut we may assume whenever multiplication looks convenient. When

P(B)>0

, dividing both sides by

P(B)

gives the equivalent statement

P(A\mid B)=P(A)

. If

P(A)>0

, the reversed condition also satisfies

P(B\mid A)=P(B)

The product definition works even if one event has probability zero. The conditional versions require their respective denominators to be positive.

In the study-club table,

P(A\mid B)=0.6

but

P(A)=0.45

. Morning attendance and bringing a laptop are therefore dependent in this hypothetical population. This comparison does not establish that attending in the morning causes someone to bring a laptop.

An independent example comes from two independent fair coin tosses. Let

A

mean the first toss is heads and

B

mean the second toss is heads. The four outcomes HH, HT, TH, TT are equally likely. Thus

P(A)=P(B)=1/2

and

P(A\cap B)=1/4=(1/2)(1/2)

. Both events can happen together, yet they are independent.

Independent is different from mutually exclusive

Mutually exclusive, or disjoint, events cannot happen together. Their intersection is empty, so its probability is zero.

On one fair die roll, let

C

be “even” and

D

be “odd.” These events are disjoint. But knowing the roll is odd makes the chance of even zero, whereas it was

1/2

before that information. They are dependent.

In general, if disjoint events each have positive probability, then

P(C\cap D)=0

while

P(C)P(D)>0

. They cannot be independent. The positive-probability qualification is essential.

A useful distinction: disjointness concerns overlap of events; independence concerns whether probability changes under information. “They have nothing to do with each other” is too vague to prove either.

Repeated draws: keep track of what remains

Rearranging the conditional definition gives the multiplication rule:

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

When using this form,

P(A)>0

is needed to define the displayed conditional factor. If

P(A)=0

, the joint probability is already zero. If

P(B)>0

, the alternative form

P(B)P(A\mid B)

is available. For independent events these reduce to

P(A)P(B)

Consider a bag of 3 red and 2 blue chips. Draw twice, uniformly at each draw, without replacement. Let

R_1

and

R_2

mean red on the first and second draws. The subscripts label positions in the sequence.

For two reds:

Before the first draw, 3 of 5 chips are red: $P(R_1)=3/5$ .
Given that the first was red, 2 red chips remain among 4: $P(R_2\mid R_1)=2/4$ .
Both must occur, so multiply along that path: $P(R_1\cap R_2)=(3/5)(2/4)=3/10$ .

The second denominator becomes 4 because one chip has left the bag. The second numerator becomes 2 because the known first chip was red.

For two of the same colour, either the red-red path or the blue-blue path qualifies:

P(\text{same colour})=\frac35\frac24+\frac25\frac14=\frac{3}{10}+\frac{1}{10}=\frac25.

We add because those complete paths are disjoint. We multiply within each path because both sequential events must happen. “And means multiply, or means add” is incomplete advice: the factors must be conditional where appropriate, and overlapping alternatives need an overlap correction.

With replacement and an independent uniform redraw, the red-red probability would instead be

(3/5)(3/5)=9/25

. State the sampling mechanism before choosing the formula.

A tree written as a table

A tree records possible paths. The table below contains the same information and remains readable without a picture.

First draw	Second given first	Path probability
Red: $3/5$	Red: $2/4$	$6/20$
Red: $3/5$	Blue: $2/4$	$6/20$
Blue: $2/5$	Red: $3/4$	$6/20$
Blue: $2/5$	Blue: $1/4$	$2/20$

The first-draw probabilities add to 1. Within each first-draw branch, the second-draw probabilities also add to 1. Finally, the four complete path probabilities total 1. These checks catch incomplete trees and incorrect remaining counts.

Total probability: combine all ways the event can happen

In the chip example, the second chip can be red after a first red or after a first blue. Therefore

P(R_2)=\frac35\frac24+\frac25\frac34=\frac{12}{20}=\frac35.

This equals the first-draw red probability, but does not make the draws independent. The marginal probability

P(R_2)=3/5

differs from the conditional probability

P(R_2\mid R_1)=1/2

. The word marginal refers to the overall probability of one event, without restricting to a particular result of the other draw.

More generally, a partition

B_1,\ldots,B_k

splits the sample space into disjoint groups that cover every outcome. The index

i

labels a group;

k

is the number of groups. If each group has positive probability,

P(A)=\sum_{i=1}^{k}P(A\mid B_i)P(B_i).

The summation symbol

\sum

says to add one term for each group. Each term is “chance of entering the group” times “chance of A within that group.” Zero-probability groups contribute no probability mass and can be omitted rather than evaluating an undefined conditional ratio.

For a hypothetical factory, machine 1 produces 60% of items with a 5% defect rate; machine 2 produces the remaining 40% with an 8% defect rate. Assuming each item comes from exactly one machine, a randomly selected item is defective with probability

0.60(0.05)+0.40(0.08)=0.062=6.2\%.

An unweighted average of 5% and 8% would incorrectly give the machines equal shares. To reverse the question and ask which machine produced a defective item, continue to Bayes' theorem.

Question-pattern anatomy

These are useful families of questions, not an exhaustive catalogue of possible exam questions. Identify the reasoning task before doing arithmetic.

Question cue	First move	Typical mistake
“Given,” “among,” or “of those who”	Name the restricted group	Divide by everyone
A table with two categories	Mark the condition's row or column	Reverse the condition
Union is given, joint is needed	Recover the overlap by addition rule	Use union as intersection
“Are these independent?”	Compare joint with product of marginals	Assume no overlap means independence
Repeated draws	State replacement and remaining counts	Keep the original denominator
Several routes to one result	Enumerate disjoint paths, then weight and add	Average conditional rates equally

Before submitting any answer, check: Does the condition have positive probability? Are my numerator and denominator measured over the same experiment? Is the result in

[0,1]

? Did I assume independence without justification?

Practice on your own

These checks use new hypothetical examples. Commit an answer before revealing its explanation. For each miss, identify whether the error was the condition, overlap, sampling mechanism, or arithmetic, then return to the corresponding explanation above.

Question 1

Type: MCQ | Marks: 1 | Topic: Choose the denominator

A club has 50 members. Of the 20 who cycle to meetings, 8 bring lunch. Altogether 15 members bring lunch. One member is selected uniformly. Given that the member cycles, what is the probability they bring lunch?

MCQ

Choose the denominator

3 Unit Assessment

8/50

8/20

8/15

15/50

Question 2

Type: NAT | Marks: 1 | Topic: Recover the overlap

In a hypothetical model,

P(A)=0.5

P(B)=0.4

, and

P(A\cup B)=0.7

. Find

P(A\mid B)

. Enter a decimal.

NAT

3 Unit Assessment

Question 3

Type: MCQ | Marks: 1 | Topic: Independence and disjointness

Let

P(A)=0.3

P(B)=0.6

, and

P(A\cap B)=0.18

. Which statement follows?

MCQ

Independence and disjointness

3 Unit Assessment

The events are disjoint and independent.

The events are disjoint but dependent.

The events are independent but not disjoint.

The events are dependent and not disjoint.

Question 4

Type: NAT | Marks: 1 | Topic: Update the bag

A bag contains 4 green and 2 yellow tokens. Two are drawn uniformly without replacement. Find the probability that both are yellow. Enter a decimal rounded to exactly four decimal places.

NAT

3 Unit Assessment

Question 5

Type: NAT | Marks: 1 | Topic: Weight the routes

A hypothetical dispatch centre routes 25% of parcels through line A and 75% through line B. The damage rates are 4% on A and 8% on B. What is the probability that a randomly selected parcel is damaged? Enter a decimal probability, not a percentage.

NAT

3 Unit Assessment

Question 6

Type: MCQ | Marks: 1 | Topic: Marginal is not conditional

A bag contains 2 red and 2 blue tokens. Two tokens are drawn uniformly without replacement. Which pair is correct: probability the second token is red; probability the second is red given that the first is red?

MCQ

Marginal is not conditional

3 Unit Assessment

1/2;\ 1/2

1/3;\ 1/2

1/2;\ 1/3

1/3;\ 1/3

Explain it without the formulas

Close the worked examples and answer in your own words:

Why does conditioning change the denominator? Use a group you can actually count.
Can two independent events happen together? Give a counterexample to “independent means disjoint.”
Why did we multiply within a chip-draw path and add between complete paths?
How can the second draw have the same red probability as the first while the draws are dependent?

Check your explanation by inventing a small count table and verifying it numerically. If you can only repeat the formula, revisit the study-club example and label the two groups aloud.

What to do next

If the denominator still feels uncertain, revisit contingency tables and retry Question 1 with a table you draw yourself. If the reasoning is solid, continue to Bayes' theorem, which reverses a condition using the same joint probability. Later, discrete random variables extends probability from named events to numerical outcomes.

For a short later review, hide this chapter and reconstruct one conditional calculation and one independence counterexample. Adjust the next review to what you can actually retrieve; a fixed interval is not a guarantee of memory. Before an imminent exam, use a short break and a fresh variant rather than repeating only the question you now recognise.

To discuss a mistake with a tutor, copy the question, your attempt, and this request: “Explain which group my denominator counts and where my reasoning first breaks. Give a hint before the answer, then an original nearby problem.” Preserve the LaTeX when copying. There are no essential images in this chapter, so the table and text carry the full explanation.

Sources and revision

Original import: the source URL and export timestamp in the front matter. Reworked by GPT-6 / Codex on 12 September 2026 as student-authored instruction. The original version is recoverable in Git. The recorded week label is inherited, not independently verified as an official assessment boundary.

Concepts cross-checked against OpenStax, Independent and Mutually Exclusive Events on 12 September 2026. Examples and six interactive checks above are hypothetical teaching material, not quoted from that textbook or derived from a graded answer key. Revision record: docs/sources/2026-09-12-foundation-learning-paths.md.