AP Statistics Topic 2.7: Independent Events and Unions of Events

01 · Start here

The concept in 60 seconds

Independent events keep each other’s probabilities unchanged. If learning that $B$ happened leaves the probability of $A$ unchanged, $A$ and $B$ are independent.

A union means $A$ or $B$ or both. Include every outcome belonging to $\text{at least }1$ event. A shared outcome belongs to the union $1\text{ time}$ , even though it belongs to both events.

For $1$ uniform draw from tickets numbered $1\text{-}8$ , let $A$ be “even” and $B$ be “greater than $4$ .” $\frac12\text{ of all }8$ numbers are even. $\frac12\text{ of the }4$ numbers greater than $4$ are also even. Here, learning $B$ does not change the chance of $A$ .

Two different questions:
Independence asks whether the information changes a probability.
A union asks which outcomes satisfy at least $1\text{ of the}$ two events.

1

draw, two events: find the shared outcomes

$1$ Neither $2$

A

only $3$ Neither $4$

A

only $5$

B

only $6$ Both $7$

B

only $8$ Both

A

· even $4$ of $8$ tickets

$A = \{2, 4, 6, 8\}$
$P(A) = 0.50$

B

· greater than

4

$4$ of $8$ tickets

$B = \{5, 6, 7, 8\}$
$P(B) = 0.50$

Each card is $1$ equally likely ticket. Dark cards satisfy both events; light olive cards satisfy $\text{exactly }1$ ; gray cards satisfy neither. Membership labels provide the same information as color.

Quick check: do independent events have to be separate?

No. The ticket events are independent and share outcomes $6$ and $8$ . Independence concerns an unchanged probability; it does not mean the events cannot occur together.

02 · Define the events clearly

The ticket model

A bag contains $8$ identical, well-mixed tickets numbered $1\text{-}8$ , $1\text{ of each}$ . $1$ ticket is drawn without looking, so all $8$ results are equally likely. This $1$ draw is $1$ complete trial.

$A$ : the number is even, so $A = \{2, 4, 6, 8\}$ .
$B$ : the number is greater than $4$ , so $B = \{5, 6, 7, 8\}$ .
$A \cap B$ : the number is even and greater than $4$ , so $A \cap B = \{6, 8\}$ .
$A \cup B$ : the number is even or greater than $4$ or both, so $A \cup B = \{2, 4, 5, 6, 7, 8\}$ .

For this lesson, $B$ means greater than $4$ . Always use the event definition in the current question, even if an earlier lesson used the same letter differently.

The probability of $A$ is $\frac{4}{8} = 0.50$ . Given $B$ , the eligible group is $\{5, 6, 7, 8\}$ , and $2\text{ of its }4$ numbers are even: $P(A\mid B) = \frac{2}{4} = 0.50$ .

The union has $6$ outcomes, so $P(A\cup B) = \frac{6}{8} = 0.75$ . Numbers $6$ and $8$ satisfy both descriptions, but each is still just $1$ ticket.

03 · Translate the question

Key ideas and notation

Intersection: $A \cap B$

$A$ and $B$ both occur.

Even and greater than $4$ : $\{6, 8\}$ .

Union: $A \cup B$

$A$ or $B$ or both occur.

Even or greater than $4$ : $\{2, 4, 5, 6, 7, 8\}$ .

Independent events

Knowing whether one occurs does not change the probability of the other.

For these ticket events, $P(A\mid B) = P(A) = 0.50$ .

Mutually exclusive events

They cannot occur together, so they have no shared outcomes.

On $1$ die roll, “result $2$ ” and “result $5$ ” are mutually exclusive.

Keep these three rules connected

General multiplication:
$P(A\cap B) = P(A) \times P(B\mid A)$ , when $P(A) > 0$ .

For independent events:
$P(A\cap B) = P(A) \times P(B)$ .

General addition:
$P(A\cup B) = P(A) + P(B) – P(A\cap B)$ .

Independence lets you replace the conditional factor with the overall probability. The addition rule works for any two events; it does not require independence.

Read “or” inclusively. Unless the question says “ $\text{exactly }1$ ,” outcomes where both events occur count toward “ $A$ or $B$ .” “ $\text{At least }1$ ” also includes both.

04 · Compare probabilities

Check independence

Method 1: compare conditional and overall probability

$P(A\mid B) = P(A)$ , provided $P(B) > 0$ .

Equivalently, compare $P(B\mid A)$ with $P(B)$ when $P(A) > 0$ .

For the ticket model, $P(A\mid B) = \frac{2}{4} = 0.50$ and $P(A) = \frac{4}{8} = 0.50$ . The values are equal, so $A$ and $B$ are independent.

Independence: the even share stays at

50\%

Before using

B

· all ticketsEven: $4$ of $8$

Even numbers: $2$ , $4$ , $6$ , $8$

$P(A) = \frac{4}{8} = 0.50$

Given

B

· greater than

4

Even: $2$ of $4$

Eligible: $5$ , $6$ , $7$ , $8$
Even within $B$ : $6$ , $8$

$P(A\mid B) = \frac{2}{4} = 0.50$

Each track represents $0$ to $100\%$ of its own reference group. Both olive sections occupy $50\%$ . The groups have different sizes, but their even proportions are equal.

Method 2: compare the joint probability with the product

$A$ and $B$ are independent if and only if $P(A\cap B) = P(A) \times P(B)$ .

In the ticket model, the actual joint probability is $\frac{2}{8} = 0.25$ . The product is $(\frac{4}{8}) \times (\frac{4}{8}) = 0.25$ . Equality confirms independence.

If the exact values differ, the events are dependent in that probability model. You need only one valid equality check; you do not have to perform both methods.

Independent does not mean mutually exclusive

If $A$ and $B$ are mutually exclusive and both have positive probability, $P(A\cap B) = 0$ while $P(A) \times P(B) > 0$ . The equality for independence fails. Learning that $A$ occurred makes $B$ impossible, changing its positive probability to $0$ .

Three event relationships: compare joint with product

Ticket modelIndependent + overlapping

Even and greater than $4$

Joint: $0.25$
Product: $0.50 \times 0.50 = 0.25$

Equal → independent.
Shared tickets $6$ and $8$ → overlap.

1\text{ fair die roll}

Mutually exclusive + dependent

Even and odd

Joint: $0$
Product: $0.50 \times 0.50 = 0.25$

Unequal → dependent.
No shared outcomes → disjoint.

60

-student club modelDependent + overlapping

Chess and Music

Joint: $0.20$
Product: $0.50 \times 0.35 = 0.175$

Unequal → dependent.
$12$ shared students → overlap.

These are different original teaching models. For the club example, probabilities describe $1\text{ uniform selection}$ from the $60$ listed students; its full table appears in Example 3.

Overlap alone also does not prove independence. Use a conditional or product comparison to decide. The positive-probability condition matters: an event with probability $0$ can be independent of another event by the product criterion. In ordinary finite equally likely models, the empty event is an example.

Exact models versus collected data: our ticket probabilities are exact. Small differences in sample percentages can occur by chance; they do not by themselves prove a relationship in a larger population. Use unrounded values when checking an exact model.

05 · Count each outcome once

Calculate a union

Adding $P(A)$ and $P(B)$ counts every shared outcome $2\text{ times}$ : $1\text{ time}$ as part of $A$ and $1\text{ time}$ as part of $B$ . Subtract the intersection $1\text{ time}$ to keep each shared outcome $1\text{ time}$ .

$P(A\cup B) = P(A) + P(B) – P(A\cap B)$ .

For the ticket model: $0.50 + 0.50 – 0.25 = 0.75$ .

The union includes

3

nonoverlapping regions

Left to right: $A$ only · Both · $B$ only · Neither. Each region has $2$ of $8$ tickets, or $25\%$ .

Included in

A \cup B

$A$ only: $2$ , $4$

Probability: $\frac{2}{8} = 0.25$

Included in

A \cup B

Both: $6$ , $8$

Probability: $\frac{2}{8} = 0.25$
Count each shared ticket $1\text{ time}$ .

Included in

A \cup B

$B$ only: $5$ , $7$

Probability: $\frac{2}{8} = 0.25$

Outside

A \cup B

Neither: $1$ , $3$

Probability: $\frac{2}{8} = 0.25$

Union: $0.25 + 0.25 + 0.25 = 0.75$

Equivalent calculation: $0.50 + 0.50 – 0.25 = 0.75$ .

The $4$ regions partition the entire $8\text{-ticket}$ sample space. Track lengths show their exact probabilities. The union consists of the first $3$ regions; the $4\text{th is its complement}$ .

When the events are independent

First calculate their intersection with the product rule, then use the addition rule:

$P(A\cup B) = P(A) + P(B) – P(A) \times P(B)$ , for independent $A$ and $B$ .

Independence does not remove the overlap. In our ticket example, the two independent events overlap with probability $0.25$ .

When the events are mutually exclusive

The intersection probability is $0$ , so the general rule simplifies to $P(A\cup B) = P(A) + P(B)$ . Use that shortcut only when the shared probability is $0$ .

Neither and $\text{exactly }1$

Neither is the complement of the union: $P(A^{c}\cap B^{c}) = 1 – P(A\cup B)$ . In the ticket model, it is $1 – 0.75 = 0.25$ , corresponding to tickets $1$ and $3$ .

$\text{Exactly }1$ includes $A$ only and $B$ only, while excluding the overlap. In the ticket model, those outcomes are $\{2, 4, 5, 7\}$ , giving $\frac{4}{8} = 0.50$ . This differs from the union’s $0.75$ .

If using a formula, $P(\text{exactly }1) = P(A) + P(B) – 2P(A\cap B)$ . You can also add the $2\text{ nonoverlapping “only” regions}$ directly.

06 · Let the event relationship guide the calculation

Choose a probability rule

Start by writing the event you want: an intersection for “and,” a union for “or,” or a conditional probability for “given.” Then check which event relationship is actually stated or established.

Choose the target, then use the event relationship

Question says ANDTarget: $A \cap B$

Independent?
Use $P(A) \times P(B)$ .

Independence not established?
Use $P(A) \times P(B\mid A)$ , when $P(A) > 0$ , or obtain the joint probability directly from the model.

Question says ORTarget: $A \cup B$

Always start with:
$P(A) + P(B) – P(A\cap B)$ .

Independent? Find the intersection using the product.

Mutually exclusive? The intersection probability is $0$ .

If the question says “given,” use a conditional probability instead. If the model does not supply enough information to find the intersection or a needed conditional probability, do not invent independence.

Replacement can change independence

Use the bag from Topic 2.6: $3$ Green tokens and $2$ Gold tokens, identical in size and well mixed. $2$ tokens are drawn sequentially, each remaining token equally likely. Let $G_1$ mean “first Green” and $G_2$ mean “second Green.”

After a first Green: replacement changes the second chance

Initial bag: $3\text{ Green}+2\text{ Gold}=5\text{ tokens}$

GreenGreenGreenGoldGold

With replacement + mixingBag returns to $3\text{ Green}+2\text{ Gold}$

GreenGreenGreenGoldGold

$P(G_2\mid G_1) = \frac{3}{5} = 0.60$
$P(G_2) = \frac{3}{5} = 0.60$

Equal → Green events independent.

$P(\text{both Green}) = (\frac{3}{5}) \times (\frac{3}{5}) = 0.36$

Without replacement $2\text{ Green}+2\text{ Gold remain}$

GreenGreenGoldGold

$P(G_2\mid G_1) = \frac{2}{4} = 0.50$
$P(G_2) = \frac{3}{5} = 0.60$

Unequal → Green events dependent.

$P(\text{both Green}) = (\frac{3}{5}) \times (\frac{2}{4}) = 0.30$

Each chip represents $1$ token. $G_1$ is first Green and $G_2$ is second Green. Overall second-draw probability and conditional second-draw probability are different quantities; the latter uses information about the first result.

Without replacement, $P(G_2\mid G_1) = \frac{2}{4} = 0.50$ , whereas the overall $P(G_2) = \frac{3}{5} = 0.60$ . Each position has the same overall Green probability by symmetry, but learning the first result changes the second-draw probability. These $2$ Green events are dependent.

With replacement and thorough mixing between draws, knowing the first result leaves the second Green probability at $\frac{3}{5}$ . In this model the draws are independent.

“ $\text{At least }1$ ” can be easier through a complement

For $2$ independent attempts with the same success probability $p$ , “no successes” has probability $(1-p)^2$ . Therefore $P(\text{at least }1\text{ success})=1-(1-p)^2$ . This includes success on either attempt and success on both.

If the attempts are dependent, calculate the probability of no successes using the appropriate conditional probabilities before subtracting from $1$ . The complement rule still works; the independent product shortcut needs justification.

07 · Follow the reasoning

Worked examples

Example 1: justify independence from outcomes

Question: $1$ ticket is drawn uniformly from numbers $1\text{-}8$ . Are $A$ : “even” and $B$ : “greater than $4$ ” independent?

$P(A) = \frac{4}{8} = 0.50$ and $P(B) = \frac{4}{8} = 0.50$ .
The shared outcomes are $\{6, 8\}$ , so $P(A\cap B) = \frac{2}{8} = 0.25$ .
$P(A) \times P(B) = 0.50 \times 0.50 = 0.25$ .

Conclusion: the joint probability equals the product, so the events are independent. Equivalently, $P(A\mid B) = \frac{2}{4} = 0.50$ equals $P(A)$ . Their $2$ shared outcomes do not contradict independence.

Example 2: “and” and “or” for a coin and a die

Question: A fair coin is tossed and a fair die is rolled independently. $1$ complete trial consists of both results. Let $H$ mean “heads” and $R$ mean “die result at least $5$ .” Find $P(H\cap R)$ and $P(H\cup R)$ .

$P(H) = \frac{1}{2}$ and $P(R) = \frac{2}{6} = \frac{1}{3}$ .
Independence gives $P(H\cap R) = (\frac{1}{2}) \times (\frac{1}{3}) = \frac{1}{6}$ .
$P(H\cup R) = \frac{1}{2} + \frac{1}{3} – \frac{1}{6} = \frac{2}{3}$ .

Outcome check: the $12$ ordered coin/die results are equally likely. $2\text{ are heads}$ with a die result of $5$ or $6$ , and $8\text{ satisfy heads}$ or a result of at least $5$ . Thus the joint and union probabilities are $\frac{2}{12}$ and $\frac{8}{12}$ .

Example 3: a union without assuming independence

Question: Select $1\text{ of the}$ $60$ listed students uniformly from the original club table. Find the chance of Chess or Music membership, and check whether the two memberships are independent.

Original fictional $60$ -student club table retained from Topics 2.5 and 2.6. $C$ : Chess, $M$ : Music.
Chess member?	Music Yes	Music No	Row total
Yes	$12$	$18$	$30$
No	$9$	$21$	$30$
Total	$21$	$39$	$60$

$P(C) = \frac{30}{60} = 0.50$ ; $P(M) = \frac{21}{60} = 0.35$ ; $P(C\cap M) = \frac{12}{60} = 0.20$ .
$P(C\cup M) = 0.50 + 0.35 – 0.20 = 0.65$ .
For independence, compare $0.20$ with $0.50 \times 0.35 = 0.175$ . They differ, so the events are dependent in this selection model.

Interpretation: $39$ of the $60$ students belong to at least $1\text{ of the}$ $2$ clubs. The $12$ students in both clubs count $1\text{ time}$ . The remaining $21$ belong to neither, with probability $\frac{21}{60} = 0.35$ .

Example 4: mutually exclusive events

Question: Roll a fair die $1\text{ time}$ . Let $A$ mean “result $2$ ” and $B$ mean “result $5$ .” Find $P(A\cup B)$ and decide whether $A$ and $B$ are independent.

The results cannot occur together on the same roll, so $P(A\cap B) = 0$ .
$P(A\cup B) = \frac{1}{6} + \frac{1}{6} – 0 = \frac{1}{3}$ .
$P(A) \times P(B) = (\frac{1}{6}) \times (\frac{1}{6}) = \frac{1}{36}$ , which is not $0$ .

Conclusion: they are mutually exclusive and dependent. Given that the result is $2$ , the chance that the same roll is $5$ becomes $0$ .

Example 5: $\text{at least }1$ success

Question: $2$ attempts are independent, each with success probability $0.30$ . Find the chance of $\text{at least }1$ success.

The complement is both attempts fail.
Each failure probability is $0.70$ , so $P(\text{both fail}) = 0.70 \times 0.70 = 0.49$ .
$P(\text{at least }1\text{ success}) = 1 – 0.49 = 0.51$ .

Union check: $0.30 + 0.30 – (0.30 \times 0.30) = 0.51$ . Simply adding gives $0.60$ and counts the both-success outcome $2\text{ times}$ .

Example 6: $\text{at least }1$ Green token

Question: Draw $2\text{ times}$ without replacement from the $3\text{-Green},2\text{-Gold}$ bag. Find the chance of at least $1$ Green token. Compare with drawing with replacement and mixing.

Without replacement, the complement is Gold then Gold.
$P(\text{both Gold}) = (\frac{2}{5}) \times (\frac{1}{4}) = 0.10$ .
$P(\text{at least }1\text{ Green}) = 1 – 0.10 = 0.90$ .
With replacement, $P(\text{both Gold}) = (\frac{2}{5}) \times (\frac{2}{5}) = 0.16$ , giving $P(\text{at least }1\text{ Green}) = 0.84$ .

Why the difference? Without replacement, drawing a Gold first leaves only $1$ Gold among $4$ tokens. The second probability must account for that first result.

08 · Show why the rule applies

Explain in context

For independence, name the comparison you made and what equality or inequality means in the situation. “They seem unrelated” is not a mathematical justification.

A complete independence response: “For the uniform ticket draw, $P(\text{even and greater than 4}) = \frac{2}{8} = 0.25$ . This equals $P(\text{even}) \times P(\text{greater than 4}) = (\frac{4}{8}) \times (\frac{4}{8}) = 0.25$ . Therefore, the events are independent.”

For a union, show both individual probabilities and the shared probability. Interpret the result using “ $\text{at least }1$ ” so the inclusion of both is clear.

A complete union response: “ $P(\text{Chess or Music}) = \frac{30}{60} + \frac{21}{60} – \frac{12}{60} = \frac{39}{60} = 0.65$ . The uniformly selected student has a $65\%$ chance of belonging to at least $1$ club, including students who belong to both.”

Improve this answer: “Add them because they are independent.”

Independence does not justify adding without correcting for overlap. For independent $A$ and $B$ , first calculate $P(A\cap B) = P(A) \times P(B)$ , then use $P(A\cup B) = P(A) + P(B) – P(A\cap B)$ . Adding alone is valid for mutually exclusive events because their shared probability is $0$ .

Before finishing: define the events, show the relationship used, substitute the probabilities and interpret the answer as an “and,” “or” or conditional probability.

09 · Find and fix mistakes

Find and fix mistakes

Common errors and the reasoning that fixes them.
Mistake	Better reasoning
Treat independent as another word for mutually exclusive.	Check whether a probability stays unchanged. Disjoint positive-probability events are dependent.
Assume overlap proves independence.	Compare the actual joint probability with the product, or compare conditional and overall probabilities.
Use equal individual probabilities as evidence of independence.	Equal $P(A)$ and $P(B)$ do not establish $P(A\cap B) = P(A) \times P(B)$ .
Multiply unconditional probabilities for every “and” question.	Use the independent product only with justified independence; otherwise use a conditional factor.
Add without subtracting overlap because events are independent.	Independent events can overlap. Subtract their joint probability $1\text{ time}$ for the union.
Treat “or” as $\text{exactly }1$ .	The union includes both. $\text{Exactly }1$ excludes the intersection.
Find “neither” by subtracting the intersection from $1$ .	Neither is the complement of the union, not the intersection.
Multiply unchanged draw probabilities without replacement.	Update the second probability after the first result.
Claim independence from event names or different activities.	Use a stated model assumption or a numerical probability comparison.
Accept impossible input probabilities or cap an answer at $1$ .	Check intersection and union bounds. An impossible result can signal incompatible inputs.

Quick checks: a union probability must be at least as large as either individual probability and no greater than $1$ . An intersection probability cannot exceed either individual probability. For example, a proposed $P(A\cap B)$ larger than $P(B)$ is impossible.

10 · Try before you reveal

Practice with hints and solutions

Write the target event and choose a rule before calculating. For an independence claim, show a numerical comparison rather than relying on the event names.

1. Independent ticket events

$1$ ticket is drawn uniformly from numbers $1\text{-}8$ . Let $A$ be “even” and $B$ be “greater than $4$ .” Check independence using the joint probability and the product.

Hint for question 1

Find the shared outcomes, then compare their probability with $(\frac{4}{8}) \times (\frac{4}{8})$ .

Solution for question 1

$A \cap B = \{6, 8\}$ , so $P(A\cap B) = \frac{2}{8} = 0.25$ . $P(A) \times P(B) = (\frac{4}{8}) \times (\frac{4}{8}) = 0.25$ . The equality shows that the events are independent.

2. A union of ticket events

For the same draw, list $A \cup B$ and find its probability. Explain why $\frac{4}{8} + \frac{4}{8}$ is too large.

Hint for question 2

The shared tickets $6$ and $8$ appear in both lists. Keep each only $1\text{ time}$ .

Solution for question 2

$A \cup B = \{2, 4, 5, 6, 7, 8\}$ , giving $\frac{6}{8} = 0.75$ . The sum $\frac{4}{8} + \frac{4}{8}$ counts the $2$ shared tickets $2\text{ times}$ . Subtract $\frac{2}{8}$ $1\text{ time}$ .

3. “Only” and “ $\text{exactly }1$ ”

For the same ticket model, find (i) $P(A\cap B^{c})$ and (ii) $P(\text{exactly }1\text{ of }A\text{ and }B)$ . Compare the second answer with $P(A\cup B)$ .

Hint for question 3

$A$ only means even but not greater than $4$ . $\text{Exactly }1$ also includes $B$ only, but excludes both.

Solution for question 3

(i) $A$ only: $\{2, 4\}$ , giving $\frac{2}{8} = 0.25$ . (ii) $\text{Exactly }1$ : $\{2, 4, 5, 7\}$ , giving $\frac{4}{8} = 0.50$ . The union is $0.75$ because it also includes the both-event outcomes $6$ and $8$ .

4. Joint, union and neither

Independent events $A$ and $B$ have probabilities $0.40$ and $0.30$ . Find $P(A\cap B)$ , $P(A\cup B)$ and $P(\text{neither})$ .

Hint for question 4

Find the intersection first. Subtract it $1\text{ time}$ for the union, then complement the union.

Solution for question 4

$P(A\cap B) = 0.40 \times 0.30 = 0.12$ . $P(A\cup B) = 0.40 + 0.30 – 0.12 = 0.58$ . $P(\text{neither}) = 1 – 0.58 = 0.42$ .

5. Use the stated joint probability

An exact probability model gives $P(A) = 0.55$ , $P(B) = 0.35$ and $P(A\cap B) = 0.20$ . Find the union and determine whether $A$ and $B$ are independent.

Hint for question 5

Use the given intersection in the addition rule. Separately compare it with $0.55 \times 0.35$ .

Solution for question 5

$P(A\cup B) = 0.55 + 0.35 – 0.20 = 0.70$ . The product is $0.1925$ , which differs from the joint probability $0.20$ . The events are dependent in this exact model.

6. Disjoint events with positive probabilities

$A$ and $B$ are mutually exclusive, with $P(A) = 0.25$ and $P(B) = 0.45$ . Find their union and decide whether they are independent.

Hint for question 6

Mutually exclusive means the joint probability is $0$ . Compare $0$ with the product.

Solution for question 6

$P(A\cup B) = 0.25 + 0.45 = 0.70$ . They are dependent: $P(A\cap B) = 0$ , while $P(A) \times P(B) = 0.1125$ . Both individual probabilities are positive.

7. Read the club table

Use the $60$ -student Chess/Music table. A student is selected uniformly. Calculate $P(\text{Music}\mid \text{Chess})$ , compare it with $P(\text{Music})$ , and find $P(\text{Chess or Music})$ .

Hint for question 7

For the conditional probability, use $30$ Chess members. For the overall and union probabilities, use all $60$ students.

Solution for question 7

$P(M\mid C) = \frac{12}{30} = 0.40$ , while $P(M) = \frac{21}{60} = 0.35$ . The values differ, so the events are dependent in this model. $P(C\cup M) = \frac{30 + 21 – 12}{60} = 0.65$ .

8. $\text{At least }1$ across $3\text{ attempts}$

$3$ attempts are mutually independent, each with success probability $0.20$ . Find the chance of $\text{at least }1$ success. Also find the chance that $\text{all }3\text{ succeed}$ .

Hint for question 8

For $\text{at least }1$ , complement the all-fail outcome. The $3$ failure probabilities can be multiplied because the attempts are independent.

Solution for question 8

$P(\text{all fail}) = 0.80^3 = 0.512$ , so $P(\text{at least }1\text{ success}) = 1 – 0.512 = 0.488$ . $P(\text{all succeed}) = 0.20^3 = 0.008$ . “ $\text{At least }1$ ” includes $1,2\text{ or }3$ successes.

9. Replacement matters

Draw $2\text{ times}$ from a bag of $3$ Green and $2$ Gold tokens, with the equal-selection and mixing assumptions stated earlier. Find $P(\text{at least }1\text{ Green})$ (i) without replacement and (ii) with replacement and mixing.

Hint for question 9

The complement is Gold then Gold. Its second-draw probability depends on replacement.

Solution for question 9

(i) Without replacement: $1 – (\frac{2}{5}) \times (\frac{1}{4}) = 0.90$ . (ii) With replacement: $1 – (\frac{2}{5}) \times (\frac{2}{5}) = 0.84$ . Use a conditional second factor for the first model.

10. Diagnose impossible inputs

Two separate proposed models give (i) $P(A) = 0.40$ , $P(B) = 0.30$ and $P(A\cap B) = 0.35$ ; (ii) $P(A) = 0.70$ , $P(B) = 0.60$ and $P(A\cap B) = 0.10$ . Explain why each set of inputs is invalid.

Hint for question 10

A shared event cannot have a probability larger than either individual event. A union cannot have a probability greater than $1$ .

Solution for question 10

(i) The intersection probability $0.35$ exceeds $P(B) = 0.30$ , which is impossible. (ii) The addition rule would give $0.70 + 0.60 – 0.10 = 1.20$ , exceeding $1$ . These are incompatible inputs, not valid probabilities to round or cap at $1$ .

11 · Keep the essentials

Quick revision

IndependenceKnowing one event leaves the other event’s probability unchanged.

Check a conditional probability

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

, with

P(B) > 0

Check a product

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

if and only if the events are independent.

General “and” rule

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

, when the conditioning event has positive probability.

General “or” rule

P(A\cup B) = P(A) + P(B) – P(A\cap B)

Mutually exclusiveThe shared probability is

0

. Positive-probability disjoint events are dependent.

NeitherSubtract the union probability from

1

$\text{At least }1$ Complement “none.” Multiply unconditional factors only when independence applies.

Questions students often ask

Can two events from the same draw be independent?

Yes. The even and greater-than- $4$ ticket events both describe $1$ draw, and their joint probability equals the product. Independence is a probability relationship; it does not require separate physical experiments.

Do equal individual probabilities prove independence?

No. On a fair die, even and odd each have probability $0.50$ , but they are mutually exclusive and dependent. Compare the joint probability with the product, or compare a valid conditional probability with its overall counterpart.

Do I need independence for the addition rule?

No. $P(A\cup B) = P(A) + P(B) – P(A\cap B)$ works for any two events. Independence helps calculate the intersection when you do not already know it.

What if rounded probabilities nearly satisfy the independence check?

Use exact fractions or unrounded values when available. If the supplied numbers are rounded, a small mismatch may reflect rounding. If the numbers are measured sample percentages, a mismatch can also reflect sampling variation. Do not claim exact population dependence from that comparison alone.

12 · Show that you understand

Final understanding check

A new original teaching example lists $100$ students and their Coding and Drama club memberships. $1$ listed student is selected uniformly at random. Let $C$ mean “Coding member” and $D$ mean “Drama member.” These counts are separate from the earlier Chess/Music example.

New original $100$ -student Coding/Drama example. These counts are separate from the Chess/Music table.
Coding member?	Drama Yes	Drama No	Row total
Yes	$12$	$28$	$40$
No	$18$	$42$	$60$
Total	$30$	$70$	$100$

Find $P(C)$ , $P(D)$ and $P(C\cap D)$ .
Check independence using the product criterion.
Calculate $P(C\mid D)$ and compare it with $P(C)$ .
Find $P(C\cup D)$ and interpret it.
Find $P(\text{neither club})$ .
Find $P(\text{exactly }1\text{ club})$ .
Explain whether the events are mutually exclusive.
In a separate model, $2$ attempts are independent and each has success probability $0.40$ . Find the probability of $\text{at least }1$ success.

Open the complete final-check solution

$P(C) = \frac{40}{100} = 0.40$ ; $P(D) = \frac{30}{100} = 0.30$ ; $P(C\cap D) = \frac{12}{100} = 0.12$ .
$P(C) \times P(D) = 0.40 \times 0.30 = 0.12$ , equal to the actual joint probability. The events are independent in this uniform-selection model.
$P(C\mid D) = \frac{12}{30} = 0.40$ , which equals $P(C)$ .
$P(C\cup D) = 0.40 + 0.30 – 0.12 = 0.58$ . There is a $58\%$ chance the selected student belongs to at least $1$ club, including students in both.
$P(\text{neither}) = \frac{42}{100} = 0.42$ , also $1 – 0.58$ .
$P(\text{exactly }1) = \frac{28 + 18}{100} = 0.46$ . The $12$ students in both are excluded.
They are not mutually exclusive, because $12$ students belong to both. Independence allows this overlap.
$P(\text{at least }1\text{ success}) = 1 – (0.60) \times (0.60) = 0.64$ .

Ready to move on? You should be able to justify independence, keep “or” inclusive, correct for overlap and distinguish a union from $\text{exactly }1$ . If you added without subtracting the shared part, review the union visual before continuing.

13 · Keep building

Continue learning

Up next · Topic 2.8Topic 2.8: Introduction to Random Variables and Probability Distributions →

Assign numerical values to random outcomes and build a distribution showing the probability of each possible value.

← Topic 2.6: Conditional Probability

Review independence checks · Review the un[Truncated]

The concept in 60 seconds

The ticket model

Key ideas and notation

Intersection: A∩BA \cap B

Union: A∪BA \cup B

Independent events

Mutually exclusive events

Keep these three rules connected

Check independence

Method 1: compare conditional and overall probability

Method 2: compare the joint probability with the product

Independent does not mean mutually exclusive

Calculate a union

When the events are independent

When the events are mutually exclusive

Neither and exactly 1\text{exactly }1

Choose a probability rule

Replacement can change independence

“At least 1\text{At least }1” can be easier through a complement

Worked examples

Example 1: justify independence from outcomes

Example 2: “and” and “or” for a coin and a die

Example 3: a union without assuming independence

Example 4: mutually exclusive events

Example 5: at least 1\text{at least }1 success

Example 6: at least 1\text{at least }1 Green token

Explain in context

Find and fix mistakes

Practice with hints and solutions

1. Independent ticket events

2. A union of ticket events

3. “Only” and “exactly 1\text{exactly }1”

4. Joint, union and neither

5. Use the stated joint probability

6. Disjoint events with positive probabilities

7. Read the club table

8. At least 1\text{At least }1 across 3 attempts3\text{ attempts}

9. Replacement matters

10. Diagnose impossible inputs

Quick revision

Questions students often ask

Final understanding check

Continue learning

Intersection: $A \cap B$

Union: $A \cup B$

Neither and $\text{exactly }1$

“ $\text{At least }1$ ” can be easier through a complement

Example 5: $\text{at least }1$ success

Example 6: $\text{at least }1$ Green token

3. “Only” and “ $\text{exactly }1$ ”

8. $\text{At least }1$ across $3\text{ attempts}$