AP Statistics Topic 2.5: Mutually Exclusive Events

01 · Start here

The concept in 60 seconds

Two events are mutually exclusive if they cannot both happen in the same trial. Another word for mutually exclusive is disjoint.

For $1\text{ numbered ticket draw}$ , “the number is even” and “the number is odd” cannot both be true. Each event can happen on its own, but there is no result that satisfies both.

The key evidence:
the joint probability—the chance that both events occur—is $0$ for mutually exclusive events.

1

ticket draw: even and odd have no shared outcome

Event

A

Even

2

4

6

8

$4$ of $8$ outcomes · $P(A) = \frac{1}{2}$

Event

B

Odd

1

3

5

7

$4$ of $8$ outcomes · $P(B) = \frac{1}{2}$

Both even and odd: no outcomes
$P(A\cap B)=\frac08=0\implies\text{mutually exclusive}$

The two event groups are separate for the $\text{same }1\text{ draw}$ . Each event has a positive individual probability, but their $\text{joint probability is }0$ .

The conclusion concerns the pair of events. It does not mean that either event is individually impossible.

Quick check: could even happen on

1

draw and odd on another?

Yes. Mutually exclusive means the two conditions cannot both hold on the same draw. It does not prevent an even result on $1$ trial and an odd result on a later trial.

02 · Define one trial

One draw, two conditions

A bag contains $8$ identical, well-mixed tickets numbered $1\text{-}8$ , $1\text{ of each}$ . $1$ ticket is drawn without looking. Each numbered ticket is equally likely.

$1$ trial is $1$ draw, and the sample space is $S = \{1, 2, 3, 4, 5, 6, 7, 8\}$ . Consider $3\text{ events}$ :

$A$ : the number is even, so $A = \{2, 4, 6, 8\}$ .
$B$ : the number is odd, so $B = \{1, 3, 5, 7\}$ .
$C$ : the number is at least $6$ , so $C = \{6, 7, 8\}$ .

Questions: Are $A$ and $B$ mutually exclusive? Are $A$ and $C$ mutually exclusive? Use shared outcomes to support each answer.

$A$ and $B$ share no ticket numbers. $A$ and $C$ share $6$ and $8$ . Changing one event changes the relationship, even though the chance process stays the same.

03 · Give each symbol a meaning

Key ideas and notation

Intersection: $A \cap B$

The set of outcomes that belong to both $A$ and $B$ .

Read $A \cap B$ as “ $A$ and $B$ .” Both conditions must hold.

Joint probability

The probability that both events occur in the same trial.

Write $P(A \cap B)$ . For the ticket events, $P(A \cap C) = \frac{2}{8}$ .

Mutually exclusive

The two events cannot occur together.

Even and odd have no shared result on $1$ draw.

Empty intersection: $\varnothing$

The symbol $\varnothing$ means there are no outcomes in the intersection.

$A \cap B = \varnothing$ for even versus odd.

Mutually exclusive events: $P(A \cap B) = 0$ .

Positive joint probability: $P(A \cap B) > 0$ . The events can happen together, so they are not mutually exclusive.

For a finite sample space with equally likely outcomes, calculate:

P(A\cap B)=\frac{\text{number of outcomes in }A\cap B}{\text{total number of outcomes in }S}

The denominator is the whole sample space. “ $A$ and $B$ ” does not ask you to restrict attention to only $A$ or only $B$ . That kind of restricted question appears in the next topic, Conditional Probability.

“And” also does not automatically mean multiply two event probabilities. First identify the joint outcomes or use the information supplied by the model.

04 · Look for a shared outcome

Check the overlap

Specify $1$ complete trial. State which result both event conditions refer to.
List or mark the two events. Include endpoints correctly, such as $6$ in “at least $6$ .”
Find the intersection. Keep only outcomes that satisfy both conditions.
Calculate the joint probability. Use the complete sample space and the model’s probabilities.
State the conclusion. A $0$ joint probability supports mutually exclusive; a positive joint probability rules it out.

Sort all

8

outcomes: even versus at least

6

Whole sample space

S

8

equally likely outcomes

A

onlyEven, below $6$

2

4

$2$ outcomes

Both

A

and

C

Even and at least $6$

6

8

$2$ outcomes

C

onlyAt least $6$ , not even

7

$1$ outcome

Neither eventNot even and below $6$

1

3

5

$3$ outcomes

Shared group $\{6, 8\}$ : $P(A \cap C) = \frac{2}{8} = 0.25$

Each ticket is placed in exactly $1$ membership region. The bordered “Both” group identifies the intersection. Card areas are not probability scales; the labeled counts give the probabilities.

Here $A \cap C = \{6, 8\}$ . Each ticket has probability $\frac{1}{8}$ , so $P(A \cap C) = \frac{2}{8} = 0.25$ . Because the joint probability is positive, $A$ and $C$ are not mutually exclusive.

$1\text{ possible shared outcome}$ is enough to show overlap in this finite, equally likely model. You do not need every outcome in $A$ to belong to $C$ .

Individual probabilities do not tell you the overlap

$2$ event pairs can have the same individual probabilities while their joint probabilities differ. You need information about which outcomes they share.

Same individual chances, different joint chances

Pair 1 · no overlap $D$ and $E$

$D = \{1, 2, 3\}$
$E = \{4, 5, 6\}$

Each probability: $\frac{3}{8}$
Shared outcomes: none
Joint probability: $0$

Pair 2 · overlap $D$ and $F$

$D = \{1, 2, 3\}$
$F = \{3, 4, 5\}$

Each probability: $\frac{3}{8}$
Shared outcome: ticket $3$
Joint probability: $\frac{1}{8}$

Both panels use $1$ uniform draw from tickets $1\text{-}8$ . The individual chances are identical; the event relationship changes because the shared outcomes change.

Knowing only that $P(D) = P(E) = \frac{3}{8}$ does not establish whether $D$ and $E$ are mutually exclusive. Their individual probabilities describe event sizes, not their shared outcomes.

05 · Find the shared cell

Read a joint probability from a two-way table

The following original teaching example lists $60$ students. A student can belong to both the Chess club and the Music club. Select $1\text{ of these}$ $60$ students uniformly at random.

Let $J$ mean “belongs to Chess” and $K$ mean “belongs to Music.” The intersection $J \cap K$ consists of the students whose answers are Yes to both questions.

Joint probability:

\frac{\text{shared cell count}}{\text{grand total}}

Original $60$ -student example: the highlighted cell is “Chess Yes and Music Yes.”
Chess member?	Music Yes	Music No	Row total
Yes	$12$	$18$	$30$
No	$9$	$21$	$30$
Total	$21$	$39$	$60$

Chess and Music: $\frac{12}{60} = 0.20 = 20\%$

The intersection is the Yes/Yes cell. The grand total is the number of students eligible for selection. This is an original fictional teaching table, not a reported survey.

There are $12$ students in both clubs, so $P(J \cap K) = \frac{12}{60} = 0.20$ . The joint probability is positive; Chess and Music membership are not mutually exclusive for this selection.

Do not use $\frac{12}{30}$ or $\frac{12}{21}$ . Those denominators restrict the group to a club. The question here selects from all $60$ students, so $60$ is the denominator.

The joint probability cannot be greater than either individual event probability. Here $P(J) = \frac{30}{60} = 0.50$ and $P(K) = \frac{21}{60} = 0.35$ . Their shared group is part of each club, and $0.20$ is no greater than either $0.50$ or $0.35$ .

06 · Check the meaning of “together”

Same trial and sound evidence

One trial can contain several steps

Suppose a trial consists of $2$ independent fair coin tosses. Record the ordered pattern as $\texttt{HH}$ , $\texttt{HT}$ , $\texttt{TH}$ or $\texttt{TT}$ , with $H$ for Heads and $T$ for Tails.

Let $F$ mean “first toss is Heads” and $G$ mean “second toss is Tails.” These events can both occur in $1$ complete trial: the pattern $\texttt{HT}$ satisfies both conditions. The conditions refer to different steps, but they can still happen together within the defined trial.

2\text{ tosses}

form

1

trial:

\texttt{HT}

can satisfy both events

$\texttt{HH}$ Not both

$\texttt{HT}$ Both

F

and

G

$\texttt{TH}$ Not both

$\texttt{TT}$ Not both

Cannot both happenFirst Heads and first Tails

$\{\texttt{HH}, \texttt{HT}\}$ and $\{\texttt{TH}, \texttt{TT}\}$
No shared pattern · joint probability $0$

Can both happenFirst Heads and second Tails

$\{\texttt{HH}, \texttt{HT}\}$ and $\{\texttt{HT}, \texttt{TT}\}$
Shared pattern $\texttt{HT}$ · joint probability $\frac{1}{4}$

$\texttt{H}=\text{Heads}$ ; $\texttt{T}=\text{Tails}$ . The model uses $2$ independent fair tosses, so each complete pattern has probability $\frac{1}{4}$ . The “Both” label marks the overlap, as well as the olive fill.

Mutually exclusive, complementary and independent

Complementary events are mutually exclusive and together cover the whole sample space. But mutually exclusive events need not be complements. On $1$ die roll, $\{1, 2\}$ and $\{5, 6\}$ are disjoint, yet $3$ and $4$ belong to neither event.

Independent describes a different relationship: learning whether one event happened does not change the other event’s chance. If $A$ and $B$ are mutually exclusive and each has positive probability, learning that $A$ occurred makes $B$ impossible. Therefore such events are not independent. Formal independence calculations come in Topic 2.7.

No observed overlap is not always an impossible overlap

A finite sample can contain $0$ joint outcomes even when the chance model permits them. To justify mutually exclusive events, use the event definitions, the complete finite outcome model, or the given joint probability.

0\text{ observed}

hits is different from

0\text{ model probability}

Chance model $\texttt{HT}$ is possible

Complete outcomes: $\texttt{HH}$ , $\texttt{HT}$ , $\texttt{TH}$ , $\texttt{TT}$

$F$ : first Heads; $G$ : second Tails.
$F \cap G = \{\texttt{HT}\}$ ; $P(F \cap G) = \frac{1}{4}$ .

Illustrative

8\text{-trial}

logNo $\texttt{HT}$ appears in this run

\texttt{HH}

\texttt{TH}

\texttt{TT}

\texttt{HH}

\texttt{TH}

\texttt{TT}

\texttt{HH}

\texttt{TT}

Observed joint frequency: $\frac{0}{8} = 0$

The log is a constructed example of possible trial results, not a reported random simulation. The absence of $\texttt{HT}$ in $8$ trials does not remove it from the model’s sample space.

If a small survey records no students in both clubs, that alone does not prove that students in the wider population cannot join both. Define the population and selection carefully. Selecting uniformly from a complete listed group with $0$ shared members is different from making a claim about a larger group based on a sample.

A very small positive joint probability also remains positive. Do not $\text{round it to }0$ before deciding whether the events are mutually exclusive.

07 · Follow the reasoning

Worked examples

Example 1: even versus odd

Question: $1$ ticket is drawn uniformly from numbers $1\text{-}8$ . Are $A$ : “even” and $B$ : “odd” mutually exclusive?

$A = \{2, 4, 6, 8\}$ ; $B = \{1, 3, 5, 7\}$ .
There are $0\text{ shared outcomes}$ , so $A \cap B = \varnothing$ .
$P(A \cap B) = \frac{0}{8} = 0$ .

Conclusion: $A$ and $B$ are mutually exclusive because $1\text{ drawn number}$ cannot be both even and odd.

Example 2: even and $\text{at least }6$

Question: In the same ticket draw, are $A$ : “even” and $C$ : “at least $6$ ” mutually exclusive?

$A = \{2, 4, 6, 8\}$ ; $C = \{6, 7, 8\}$ .
The shared outcomes are $\{6, 8\}$ .
$P(A \cap C) = \frac{2}{8} = 0.25$ .

Conclusion: the events are not mutually exclusive, because ticket $6$ or ticket $8$ satisfies both conditions and the joint probability is positive.

Example 3: disjoint does not mean complementary

Question: A fair die is $\text{rolled }1\text{ time}$ . Let $L$ mean “at most $2$ ” and $H$ mean “at least $5$ .” Are the events mutually exclusive? Are they complements?

$L = \{1, 2\}$ ; $H = \{5, 6\}$ .
$L \cap H = \varnothing$ , so $P(L \cap H) = 0$ . They are mutually exclusive.
Results $3$ and $4$ belong to neither event. Together $L$ and $H$ do not cover $S$ .

Conclusion: they are not complements. The complement of $L$ is $\{3, 4, 5, 6\}$ , which is larger than $H$ .

Example 4: different tosses within one trial

Question: For $2$ independent fair tosses, let $F$ mean “first toss is Heads” and $G$ mean “second toss is Tails.” Are $F$ and $G$ mutually exclusive?

$S = \{\texttt{HH}, \texttt{HT}, \texttt{TH}, \texttt{TT}\}$ , with $4$ equally likely patterns.
$F = \{\texttt{HH}, \texttt{HT}\}$ ; $G = \{\texttt{HT}, \texttt{TT}\}$ .
$F \cap G = \{\texttt{HT}\}$ , so $P(F \cap G) = \frac{1}{4} = 0.25$ .

Conclusion: they are not mutually exclusive. Both conditions hold in the complete trial $\texttt{HT}$ .

Example 5: use the whole table total

Question: $1$ student is selected uniformly from the $60$ -student club table. Justify whether Chess and Music membership are mutually exclusive.

The shared cell contains $12$ students.
The selection group contains $60$ students.
$P(J\cap K) = \frac{12}{60} = 0.20$ .

Conclusion: the memberships are not mutually exclusive, because $20\%$ of the selectable students belong to both. The numerator is the shared cell; the denominator is the grand total.

Example 6: a small overlap is still an overlap

Question: A model gives $P(R \cap T) = 0.002$ . A student rounds this to $0.00$ and calls $R$ and $T$ mutually exclusive. Is that justified?

The given joint probability is $0.002 = 0.2\%$ .
It is small, but it is greater than $0$ .
The events therefore have a possible joint occurrence under the model.

Conclusion: they are not mutually exclusive. Use the unrounded joint probability when judging whether overlap exists.

08 · Support the conclusion

Explain in context

Give the shared outcomes or shared group, calculate the joint probability, and connect that evidence to the conclusion.

When there is no overlap: “Even and odd have $0\text{ shared outcomes}$ on $1$ draw from tickets $1\text{-}8$ , so $P(A \cap B) = \frac{0}{8} = 0$ . Thus the events are mutually exclusive.”

When there is overlap: “Tickets $6$ and $8$ are both even and at least $6$ . Therefore $P(A \cap C) = \frac{2}{8} = 0.25 > 0$ , so the events are not mutually exclusive.”

“They are different events” is not enough. $2\text{ differently named conditions}$ can apply to the same outcome. “Both have positive probability” is also not enough: even and odd each have positive probability, but their $\text{joint probability is }0$ .

Improve this answer: “Chess and Music are not mutually exclusive because they are clubs.”

The type of activity does not establish overlap. Use the given evidence: “ $12\text{ of}$ the $60$ selectable students belong to both clubs, so $P(J\cap K) = \frac{12}{60} = 0.20$ . Because the joint probability is positive, these memberships are not mutually exclusive.”

An AP-style justification should contain: the event definitions, evidence about their intersection, the joint probability when available, and a clear conclusion about the pair.

09 · Find and fix mistakes

Find and fix mistakes

Common errors and the reasoning that fixes them.
Mistake	Better reasoning
Different event names must mean no overlap.	List shared outcomes. “Even” and “at least $6$ ” share tickets $6$ and $8$ .
Each event can occur, so they cannot be mutually exclusive.	Inspect the joint probability. Even and odd each can occur, but never together on $1$ draw.
“ $A$ and $B$ ” means multiply $P(A)$ by $P(B)$ automatically.	“And” names the intersection. Determine the event relationship before choosing a multiplication rule.
Use a row total as the denominator for a joint table probability.	Use the grand total for the full selection group: $\frac{12}{60}$ , not $\frac{12}{30}$ .
Disjoint events must cover the sample space.	Covering the whole space is an extra condition for complements. Disjoint events may leave outcomes outside both.
Mutually exclusive and independent mean the same thing.	“Cannot occur together” differs from “one event does not change the other’s chance.”
Different steps in a trial cannot occur together.	A complete $2\text{-toss}$ trial $\texttt{HT}$ satisfies first Heads and second Tails.
No observed shared outcomes proves no possible shared outcomes.	A finite sample can miss an outcome that is possible under the model.
Round a small positive $\text{joint probability to }0$ , then classify.	Use the given unrounded value. Any positive joint probability rules out mutually exclusive events.
$P(A) + P(B)$ being below $1$ proves the events are disjoint.	Individual event probabilities do not determine their joint probability. Use overlap information.

Fast check: a joint probability is between $0$ and $1$ and cannot exceed either event’s individual probability. If you call two events mutually exclusive, make sure your evidence actually gives a $0$ joint probability.

10 · Try before you reveal

Practice with hints and solutions

For each pair, identify the intersection before deciding. State the joint probability when the information allows it.

1. Odd and even

A fair die is $\text{rolled }1\text{ time}$ . Let $A$ mean “odd” and $B$ mean “even.” Find $A \cap B$ and $P(A \cap B)$ . Are the events mutually exclusive?

Hint for question 1

List $\{1, 3, 5\}$ and $\{2, 4, 6\}$ . Look for a number that belongs to both.

Solution for question 1

$A \cap B = \varnothing$ . $P(A \cap B) = \frac{0}{6} = 0$ . Yes, the events are mutually exclusive: $1\text{ die result}$ cannot be both odd and even.

2. Even and $\text{at least }4$

A fair die is $\text{rolled }1\text{ time}$ . Let $A$ mean “even” and $C$ mean “at least $4$ .” Calculate the joint probability and justify whether the events are mutually exclusive.

Hint for question 2

The number $4$ belongs to “at least $4$ .” Which event outcomes are even?

Solution for question 2

$A = \{2, 4, 6\}$ ; $C = \{4, 5, 6\}$ . $A \cap C = \{4, 6\}$ , so $P(A \cap C) = \frac{2}{6} = \frac{1}{3}$ . The events are not mutually exclusive because their joint probability is positive.

3. Are disjoint events always complements?

$1$ ticket is drawn uniformly from numbers $1\text{-}8$ . Let $D$ mean “at most $3$ ” and $E$ mean “at least $6$ .” Are $D$ and $E$ mutually exclusive? Are they complements?

Hint for question 3

Look for shared outcomes, then separately check whether the two events cover every ticket number.

Solution for question 3

$D = \{1, 2, 3\}$ ; $E = \{6, 7, 8\}$ . Their intersection is empty, so the joint probability is $0$ and they are mutually exclusive. They are not complements: tickets $4$ and $5$ belong to neither event.

4. First Head and second Head

$1$ trial consists of $2$ independent fair coin tosses. Let $F$ mean “first toss is Heads” and $G$ mean “second toss is Heads.” List $F \cap G$ and find its probability.

Hint for question 4

Use $\texttt{HH}$ , $\texttt{HT}$ , $\texttt{TH}$ and $\texttt{TT}$ . Both event conditions refer to the same complete $2\text{-toss}$ trial.

Solution for question 4

$F = \{\texttt{HH}, \texttt{HT}\}$ ; $G = \{\texttt{HH}, \texttt{TH}\}$ . $F \cap G = \{\texttt{HH}\}$ . $P(F \cap G) = \frac{1}{4} = 0.25$ , so the events are not mutually exclusive.

5. Not Chess and Music

Use the $60$ -student club table above. $1$ listed student is selected uniformly. Find $P(J^{c}\cap K)$ . Are these two membership conditions mutually exclusive?

Hint for question 5

Find the cell with No to Chess and Yes to Music. Keep $60$ as the denominator.

Solution for question 5

The joint cell contains $9$ students, so the probability is $\frac{9}{60} = 0.15$ . The conditions are not mutually exclusive: those $9$ students satisfy both.

6. Check the response rule

A travel survey asks each student to select exactly $1$ primary method: Bus, Walk or Cycle. A club survey allows students to select every club they belong to. Can you automatically call Bus and Cycle mutually exclusive? Can you automatically do the same for Chess and Music?

Hint for question 6

Ask whether $1$ student’s recorded response can meet both conditions under each survey’s rules.

Solution for question 6

Bus and Cycle are mutually exclusive for the recorded primary-method response, because exactly $1$ method is allowed. Chess and Music are not automatically mutually exclusive: multiple club memberships are allowed, so $1$ student could belong to both. Use actual event definitions and joint information to assess the memberships.

7. $0$ versus small positive probability

$3$ event pairs have joint probabilities $0$ , $0.04$ and $0.0005$ . Which pair is mutually exclusive? Explain why rounding can be misleading.

Hint for question 7

Compare each given probability with $\text{exactly }0$ before rounding.

Solution for question 7

Only the pair with joint probability $0$ is mutually exclusive. The other probabilities are positive: $0.04 = 4\%$ and $0.0005 = 0.05\%$ . A rounded display of $0.00$ can conceal a small positive overlap.

8. Same individual probabilities

For $1$ uniform draw from tickets $1\text{-}8$ , $D = \{1, 2, 3\}$ , $E = \{4, 5, 6\}$ , and $F = \{3, 4, 5\}$ . $\text{All }3\text{ have probability}$ $\frac{3}{8}$ . Compare $D$ with $E$ and $D$ with $F$ .

Hint for question 8

The event sizes are equal. Their intersections need not be.

Solution for question 8

$D \cap E = \varnothing$ , so $P(D \cap E) = 0$ : this pair is mutually exclusive. $D \cap F = \{3\}$ , so $P(D \cap F) = \frac{1}{8} = 0.125$ : this pair is not mutually exclusive. Individual event probabilities alone do not determine overlap.

9. No overlap in $8\text{ trials}$

Under the independent fair $2\text{-toss}$ model, $F$ means “first toss is Heads” and $G$ means “second toss is Tails.” A run of $8$ trials contains no $\texttt{HT}$ patterns. Does that make $F$ and $G$ mutually exclusive?

Hint for question 9

Use the model’s complete sample space, rather than treating the observed log as every possible outcome.

Solution for question 9

No. $\texttt{HT}$ remains a possible outcome with probability $\frac{1}{4}$ . Thus $P(F \cap G) = 0.25$ under the model. $0\text{ observed}$ $\texttt{HT}$ patterns gives a joint relative frequency of $\frac{0}{8}$ in that run, not a $0\text{ model probability}$ .

10. Unequal outcome probabilities

A model selects exactly $1$ color, with probabilities Red $0.40$ , Blue $0.25$ , Gold $0.15$ , Green $0.10$ and White $0.10$ . Let $M$ mean “Blue or Gold” and $N$ mean “Gold or White.” Find $P(M \cap N)$ and justify whether $M$ and $N$ are mutually exclusive.

Hint for question 10

Which color belongs to both event lists? Use its supplied probability, since the colors are not equally likely.

Solution for question 10

$M \cap N$ contains only Gold. $P(M \cap N) = 0.15$ , so the events are not mutually exclusive. The calculation is not $\frac{1}{5}$ : the $5$ color labels have unequal probabilities.

11 · Keep the essentials

Quick revision

Define the trialBoth conditions must refer to the same complete result.

Read $\cap$ as “and”The intersection contains outcomes satisfying both event definitions.

Find the shared outcomesUse event lists, membership regions or a shared table cell.

Use the whole denominatorA joint probability refers to the complete selection group.

Joint probability $0$ This is the probability evidence for mutually exclusive events.

Joint probability positiveThe events can occur together and are not mutually exclusive.

Check the source of a $0$ An observed

0

in a small run need not be a

0\text{ model probability}

Justify in contextName the overlap or explain why no outcome can satisfy both conditions.

Questions students often ask

Do mutually exclusive events each have

\text{probability }0

No. Their joint probability is $0$ . For $1\text{ fair die roll}$ , odd and even each have probability $\frac{1}{2}$ , but the probability of a result that is both is $0$ .

Are mutually exclusive events always complements?

No. Complements also cover the whole sample space. Two disjoint events may leave outcomes outside both, such as $\{1, 2\}$ and $\{5, 6\}$ on a die.

Does “and” tell me to multiply?

No. It names the intersection. Use shared outcomes, a joint table cell or a supplied joint probability. Multiplication rules require additional information about the event relationship.

Can different toss positions still overlap?

Yes. In a complete $2\text{-toss}$ trial, “first Heads” and “second Tails” both hold for $\texttt{HT}$ . State the trial before deciding whether the events can occur together.

12 · Show that you understand

Final understanding check

An original teaching example lists $100$ students and their Art and Science club memberships. $1\text{ of these}$ listed students is selected uniformly at random. Membership in both clubs is allowed.

Original $100$ -student membership example: $1$ listed student is selected uniformly.
Art member?	Science Yes	Science No	Row total
Yes	$12$	$28$	$40$
No	$18$	$42$	$60$
Total	$30$	$70$	$100$

Let $A$ mean “Art member” and $B$ mean “Science member.”

Describe $A \cap B$ and calculate its probability.
Justify whether $A$ and $B$ are mutually exclusive.
Are $A$ and “not $A$ ” mutually exclusive? Are they complements?
Find $P(A\cap B^{c})$ . Are $A$ and not $B$ mutually exclusive?
A student argues, “ $P(A) = 0.40$ and $P(B) = 0.30$ , and their sum is below $1$ , so the events must be disjoint.” Explain the mistake.
If a small sample from a wider school population has no students in both clubs, is that enough to prove that Art and Science memberships in the whole school are mutually exclusive?

Open the complete final-check solution

$A \cap B$ means membership in both Art and Science. $P(A \cap B) = \frac{12}{100} = 0.12$ .
Not mutually exclusive: the joint probability is positive, and $12$ selectable students satisfy both conditions.
Yes to both. $A$ and not $A$ cannot occur together and cover the entire selection group. Their joint probability is $0$ .
The Art Yes and Science No cell has $28$ students. $P(A\cap B^{c}) = \frac{28}{100} = 0.28$ . These two conditions are not mutually exclusive.
Individual probabilities do not determine the overlap. The given table actually shows a positive joint probability of $0.12$ , which rules out mutually exclusive events.
No. No observed joint members in a small sample does not establish that membership in both clubs is impossible in the whole school. Use the population’s event definitions or adequate joint information.

Ready to move on? You should be able to identify the intersection, calculate its probability, and justify the event relationship. If you used a row total for a joint probability, revisit the two-way table before continuing.

13 · Keep building

Continue learning

Up next · Topic 2.6Topic 2.6: Conditional Probability →

Restrict the selection group when information is given. Learn how a conditional probability differs from the joint probability calculated in this lesson.

← Topic 2.4: Introduction to Probability

Review shared outcomes · Review joint table probabilities · Back to the lesson overview

0000003998 : Abdad Alam Shamim Alam

This is the bestest institution I have ever come to and I love it very much.

Aliki S

Ali

Sly is him

Jitendra Kumar Kumawat

Raahil Hasan

Tina Mudarres

The classes are amazing and my child learnt so much

alisha gadoya

I have experienced a lot of good things, it has taught me so many things and from B/Cs I have gone to an A! This is wonderful and Mavish’s class is awesome.

smasher 123

It’s sensational my kid went. First he only would get C now it’s all A’s really good and recommend

Mosa Al- Samaraie

AMAZING PRICES GREAT TEACHERS STRAIGHT FORWARD LEARNING STEADY PACE IN TUTORING

Cael Dagnelie

I had a great experience learning math with Ms. Mavish. She explains complex topics in a very clear and simple way, which made it easier for me to understand and enjoy the subject. Her patience and dedication really stood out, and she always made sure that everyone in the class was keeping up. I especially appreciated how approachable she was — I never felt afraid to ask questions, and she was always willing to help. Thanks to her teaching, my confidence in math has grown a lot. I’m really thankful for the effort she puts into every lesson!