AP Statistics Topic 2.4: Introduction to Probability

01 · Start here

The concept in 60 seconds

Probability describes how likely an event is under a stated chance model. We write it as a number from $0$ to $1$ , or as a percentage from $0\%$ to $100\%$ .

A useful starting point is to list the possible outcomes. If those outcomes are equally likely, the probability of an event is the share of outcomes that satisfy its condition.

For equally likely outcomes:

P(E)=\frac{\text{number of outcomes in }E}{\text{number of outcomes in }S}

8

equally likely tickets: mark the event “at least

6

”

$1$ Not

E

$2$ Not

E

$3$ Not

E

$4$ Not

E

$5$ Not

E

$6$ In

E

$7$ In

E

$8$ In

E

$P(E)=\frac38=37.5\%$

Every numbered ticket represents $1\text{ possible outcome}$ . Text labels identify event membership as well as color.

There are $3$ eligible ticket numbers among $8$ equally likely possibilities. The probability is $\frac{3}{8} = 0.375 = 37.5\%$ . We count all $8$ possible outcomes in the denominator, including the ones outside the event.

Quick check: does a

37.5\%

chance mean the next draw must be

6

7

8

No. Any of the $8$ numbers can occur on the next draw. The probability describes the event’s chance under the model; it does not tell us which result the next draw will produce.

02 · State the model and question

An $8\text{-ticket}$ question

A bag contains $8$ identical tickets, numbered $1$ through $8$ , with $1$ ticket of each number. The tickets are mixed thoroughly, and $1\text{ is drawn}$ without looking. Our model assumes that each ticket has the same chance of being selected.

Question: What is the probability of drawing a number at least $6$ ?

$1\text{ trial}$ consists of $1$ draw. Its outcome is $1\text{ ticket number}$ . “At least $6$ ” means $6$ or larger, so the event includes $6$ , $7$ and $8$ .

The words matter. “Greater than $6$ ” would include only $7$ and $8$ . Before calculating, translate the question into an exact list of outcomes.

If we repeat this same $1\text{-draw}$ model, we return the ticket and mix the bag before each new trial. Otherwise, the bag changes and the original $8\text{-outcome}$ model no longer describes every draw.

03 · Give each word a job

Key ideas and notation

Sample space: $S$

The set of all possible, nonoverlapping outcomes for the trial.

$S = \{1, 2, 3, 4, 5, 6, 7, 8\}$ for $1$ ticket draw.

Outcome

One possible result of a trial.

Drawing ticket $7$ is $1\text{ outcome}$ . List each outcome $1\text{ time}$ .

Event: $E$

A set of outcomes that satisfy a specified condition.

$E = \{6, 7, 8\}$ means the drawn number is at least $6$ .

Complement: $E^c$

All outcomes in $S$ that are outside $E$ : the event “not $E$ .”

$E^c = \{1, 2, 3, 4, 5\}$ . The number is less than $6$ .

1

draw, four different descriptions

Trial $1$ ticket draw

Perform the random process $1\text{ time}$ , then record the number.

1\text{ outcome}

Ticket $7$

One particular result of that draw.

Event

E

A number at least $6$

A set of eligible outcomes: $\{6, 7, 8\}$ .

Sample space

S

All $8$ numbers

$\{1, 2, 3, 4, 5, 6, 7, 8\}$ : the complete list.

An event can contain several outcomes. The sample space must include every possible outcome, whether it is inside or outside the event.

Read the symbols in words

$P(E)$ means “the probability that event $E$ occurs.” $P(E^c)$ means “the probability that $E$ does not occur.” You may also see the complement written $E^{\prime}$ , $E^C$ , or $\overline{E}$ . They express the same idea here.

$0 \le P(E) \le 1$
Every event probability lies between $0$ and $1$ , inclusive.

$P(S) = 1$
One of the outcomes in the complete sample space must occur.

In our finite ticket model, “draw $9$ ” has no possible outcomes and has probability $0$ . “Draw a number from $1$ through $8$ ” covers the whole sample space and has probability $1$ .

A probability such as $1.2$ or $-0.1$ is invalid. A probability such as $0.08$ is valid and equals $8\%$ , not $0.08\%$ .

04 · List all possibilities

Build a sample space

Define $1\text{ trial}$ . Are we drawing $1$ ticket, tossing $2$ coins, or rolling $2$ dice?
Choose enough detail. Record order or object identity when it distinguishes outcomes.
List every possible outcome $1\text{ time}$ . No gaps, duplicates or overlapping cases.
Mark the outcomes in the event. Check the exact wording before counting.

$2\text{ dice}$ : keep the dice distinguishable

Roll $1\text{ red die}$ and $1\text{ blue die}$ . Assume the dice are fair and their results are independent. Record an ordered pair (red result, blue result). Then $(1, 6)$ and $(6, 1)$ are different outcomes, even though both have sum $7$ .

2\text{-dice}

grid:

36

ordered pairs,

6\text{ with sum}

7

Red ↓ Blue →	$1$	$2$	$3$	$4$	$5$	$6$
$1$	$2$	$3$	$4$	$5$	$6$	$7$ ★
$2$	$3$	$4$	$5$	$6$	$7$ ★	$8$
$3$	$4$	$5$	$6$	$7$ ★	$8$	$9$
$4$	$5$	$6$	$7$ ★	$8$	$9$	$10$
$5$	$6$	$7$ ★	$8$	$9$	$10$	$11$
$6$	$7$ ★	$8$	$9$	$10$	$11$	$12$

Read each cell by its row and column: red $2$ , blue $5$ identifies $(2, 5)$ . The number displayed is the sum. A star and an olive fill mark each sum- $7$ cell.

The event “sum $7$ ” includes $(1, 6)$ , $(2, 5)$ , $(3, 4)$ , $(4, 3)$ , $(5, 2)$ and $(6, 1)$ . These are $6$ outcomes among $36$ equally likely ordered pairs, so its probability is $\frac{6}{36} = \frac{1}{6}$ .

We could describe outcomes only by their sums, from $2$ through $12$ . That is a valid sample space for the sum, but its $11$ outcomes are not equally likely. Sum $2$ has just $1\text{ pair}$ , $(1, 1)$ , while sum $7$ has $6$ pairs. Counting the $11\text{ sum labels}$ as if each had probability $\frac{1}{11}$ would use the wrong model.

05 · Check before you count

Check equally likely outcomes

“Equally likely” means that every outcome in the chosen sample space has the same probability. A process being random does not automatically make every label equally likely.

P(E)=\frac{\text{number of outcomes in }E}{\text{number of outcomes in }S}

Condition: all outcomes counted in $S$ must be equally likely.

$2\text{ fair coin tosses: }4\text{ patterns, }3\text{ head counts}$

Toss a fair $\text{coin }2\text{ times}$ , with independent results. Use $\texttt{H}$ for Heads and $\texttt{T}$ for Tails. The ordered patterns are $\texttt{HH}$ , $\texttt{HT}$ , $\texttt{TH}$ and $\texttt{TT}$ . Each has probability $\frac{1}{4}$ under this model.

4\text{ equal pattern chances}

become unequal head-count chances

$\texttt{HH}$ Chance:

\frac{1}{4}

$\texttt{HT}$ Chance:

\frac{1}{4}

$\texttt{TH}$ Chance:

\frac{1}{4}

$\texttt{TT}$ Chance:

\frac{1}{4}

Group the same $4$ patterns by the number of Heads. $\texttt{H}=\text{Heads}$ ; $\texttt{T}=\text{Tails}$ .
Head count	Patterns in the group	Probability
$0$	$\texttt{TT}$	$\frac{1}{4} = 25\%$
$1$	$\texttt{HT}$ , $\texttt{TH}$	$\frac{2}{4} = 50\%$
$2$	$\texttt{HH}$	$\frac{1}{4} = 25\%$

The model assumes $2$ independent fair tosses. $\texttt{HT}$ and $\texttt{TH}$ are different ordered outcomes, but both contribute to the group “ $1$ Head.”

There are $3$ possible head counts— $0$ , $1$ and $2$ —but $1\text{ head}$ can happen in $2$ ways. Its probability is $\frac{2}{4} = \frac{1}{2}$ , not $\frac{1}{3}$ .

What if the outcomes are not equally likely?

Use the probabilities assigned by the model. For example, if a fictional arrival model gives $P(\text{Early})=0.15$ , $P(\text{On time})=0.75$ and $P(\text{Late})=0.10$ , the $3\text{ category labels}$ do not each have chance $\frac{1}{3}$ . A valid complete model assigns probabilities between $0$ and $1$ that total $1$ .

If no probabilities or equal-likelihood assumption are supplied, naming the possible outcomes alone may not give enough information to calculate a probability.

Connect this to simulation

In Topic 2.3, $\frac{\text{event hits}}{\text{completed trials}}$ gave a simulation estimate. Here, $\frac{\text{eligible outcomes}}{\text{all equally likely outcomes}}$ gives the model probability. A finite simulation can differ from that model value because random results vary from run to run.

06 · Calculate the chance of “not”

Find the complement

An event and its complement split the complete sample space into $2$ groups. Every outcome is in one of those groups, and no outcome is in both.

The complement rule:
$P(E^c) = 1 – P(E)$ .

Equivalently, $P(E) + P(E^c) = 1$ .

The whole chance splits into

E

and not

E

Whole sample space: probability $1 = 100\%$

Olive segment ·

E

At least $6$ : $37.5\%$

Outcomes: $\{6, 7, 8\}$
Probability: $\frac{3}{8}$

Light segment · not

E

Less than $6$ : $62.5\%$

Outcomes: $\{1, 2, 3, 4, 5\}$
Probability: $\frac{5}{8}$

The segment widths show $\frac{3}{8}$ and $\frac{5}{8}$ of the whole. Together the $2$ groups contain all $8$ tickets, with $\text{none counted }2\text{ times}$ .

For the ticket event, $P(E^c) = 1 – \frac{3}{8} = \frac{5}{8} = 0.625 = 62.5\%$ . This means the number is less than $6$ . Notice that ticket $6$ belongs to $E$ , so it cannot also belong to the complement.

The complement rule works whether or not the underlying outcomes are equally likely. It also does not require an independence assumption: it follows from $E$ and “not $E$ ” covering the whole sample space without overlap.

Negate the condition, not just a word

The complement of “at least $1$ Head” is “ $0\text{ Heads}$ .” The complement of “exactly $1$ Head” is broader: it includes every head count $\text{other than }1$ .

2

tosses: “not

\text{at least }1

” differs from “not

\text{exactly }1

”

At least

1

HeadComplement: $0\text{ Heads}$

Event outcomes

\texttt{HH}

\texttt{HT}

\texttt{TH}

Complement outcomes

\texttt{TT}

Exactly

1

HeadComplement: $0\text{ or }2$ Heads

Event outcomes

\texttt{HT}

\texttt{TH}

Complement outcomes

\texttt{HH}

\texttt{TT}

For independent fair tosses, the first complement has probability $\frac{1}{4}$ ; the second has probability $\frac{2}{4}$ . List the outcomes to check that nothing is missing.

For $1\text{ die}$ , the complement of “at most $4$ ” is “greater than $4$ ,” so it includes $5$ and $6$ . The complement of “greater than $4$ ” is “at most $4$ ,” so it includes $1$ , $2$ , $3$ and $4$ .

Keep the units consistent. With a decimal probability, calculate $1 – 0.375 = 0.625$ . With percentages, calculate $100\% – 37.5\% = 62.5\%$ .

07 · Follow the reasoning

Worked examples

Example 1: an event and its complement

Question: $1\text{ of the }8$ identical, well-mixed tickets numbered $1\text{-}8$ is drawn. Find the probability of a number at least $6$ , and the probability of its complement.

The sample space has $8$ equally likely outcomes.
$E = \{6, 7, 8\}$ , so $P(E) = \frac{3}{8} = 0.375$ .
$E^c = \{1, 2, 3, 4, 5\}$ , so $P(E^c) = 1 – 0.375 = 0.625$ .

Check: $0.375 + 0.625 = 1$ . In context, the chance of drawing a number less than $6$ is $62.5\%$ .

Example 2: watch the endpoint

Question: A fair $6\text{-sided}$ die is rolled $1\text{ time}$ . What is the probability of a result greater than $4$ ? Describe the complement.

$S = \{1, 2, 3, 4, 5, 6\}$ ; the $6$ outcomes are equally likely.
$E = \{5, 6\}$ , so $P(E) = \frac{2}{6} = \frac{1}{3}$ .
“Not greater than $4$ ” means at most $4$ : $E^c = \{1, 2, 3, 4\}$ .
$P(E^c) = 1 – \frac{1}{3} = \frac{2}{3}$ .

Why this matters: “at least $4$ ” includes $4$ , $5$ and $6$ . It is a different event and is not the complement of “greater than $4$ .”

Example 3: count ordered dice pairs

Question: $2\text{ distinguishable fair dice}$ are rolled independently. Find the probability that their sum is $7$ .

The sample space contains $36$ equally likely ordered pairs.
List the $6\text{ event outcomes}$ : $(1, 6)$ , $(2, 5)$ , $(3, 4)$ , $(4, 3)$ , $(5, 2)$ , $(6, 1)$ .
$P(\text{sum 7}) = \frac{6}{36} = \frac{1}{6} \approx 0.1667$ .

Check: sums $2\text{-}12$ are not equally likely, so “ $1\text{ desired sum out of }11$ labels” is not a valid calculation.

Example 4: “ $\text{not exactly }1$ ”

Question: $2$ independent fair coin tosses are recorded. Find the probability of exactly $1$ Head and of not exactly $1$ Head.

$S = \{\texttt{HH}, \texttt{HT}, \texttt{TH}, \texttt{TT}\}$ ; each pattern has chance $\frac{1}{4}$ .
Exactly $1$ Head gives $E = \{\texttt{HT}, \texttt{TH}\}$ , so $P(E) = \frac{2}{4} = 0.50$ .
The complement is $E^c = \{\texttt{HH}, \texttt{TT}\}$ : $2$ Heads or $0$ Heads.
$P(E^c) = 1 – 0.50 = 0.50$ .

Check: “ $0\text{ Heads}$ ” alone is just $\texttt{TT}$ , with probability $\frac{1}{4}$ . It leaves out $\texttt{HH}$ and therefore is not the full complement of exactly $1$ Head.

Example 5: unequal probabilities still have a complement

Question: A fictional arrival model assigns $P(\text{Early})=0.15$ , $P(\text{On time})=0.75$ and $P(\text{Late})=0.10$ . What is the probability of not being late?

The complete model totals $0.15 + 0.75 + 0.10 = 1$ .
The complement of Late includes Early and On time.
$P(\text{not Late}) = 1 – 0.10 = 0.90 = 90\%$ .

Check: the $2$ complement categories have probabilities $0.15 + 0.75 = 0.90$ . Counting “ $2\text{ out of }3\text{ categories}$ ” would incorrectly assume equal category probabilities.

Example 6: interpret a small chance

Question: A game model gives a bonus with probability $0.015$ on $1\text{ play}$ . Find and interpret the probability of no bonus.

The event is “receive a bonus on $1\text{ play}$ .” Its complement is “receive no bonus on that play.”
$P(\text{no bonus}) = 1 – 0.015 = 0.985 = 98.5\%$ .

Interpretation: under the model, $1\text{ play}$ has a $98.5\%$ chance of giving no bonus. A bonus is unlikely, but it can still happen on the next play. The model does not fix the number of bonuses in a particular short run.

08 · Make your answer meaningful

Explain in context

A strong response identifies the chance model, names the event, shows the calculation, and interprets the result using the objects in the question.

A complete ticket response: “There are $8$ equally likely ticket numbers. $3\text{—}6$ , $7$ and $8$ —satisfy ‘at least $6$ ,’ so the probability is $\frac{3}{8} = 0.375$ . Thus $1$ draw has a $37.5\%$ chance of giving a number at least $6$ . The complement is a number less than $6$ , with probability $1 – 0.375 = 0.625$ .”

If the ticket is replaced and mixed after every draw, the event’s relative frequency would tend toward $37.5\%$ over many repetitions of this model. A particular run of $8$ draws need not contain exactly $3$ eligible results.

Improve this answer: “There are

3

numbers, so the chance is

\frac{1}{3}

.”

The $3$ numbers are the event outcomes, not the complete sample space. $\text{The denominator is }8$ . A better answer is: “The $8$ tickets are equally likely, and $3\text{ have numbers}$ at least $6$ , so $P(E) = \frac{3}{8} = 0.375$ .”

Before finishing: ask whether your answer refers to $1\text{ trial}$ , uses the correct denominator, states the condition that justifies counting, and describes the complement with its endpoints included correctly.

09 · Find and fix mistakes

Find and fix mistakes

Common errors and the reasoning that fixes them.
Mistake	Better reasoning
“Random” means every label is equally likely.	Check the model. $2$ fair tosses give head counts $0$ , $1$ and $2$ with unequal chances.
Count the $11$ dice sums as $11\text{ equal outcomes}$ .	Use the $36$ equally likely ordered pairs, or use the correct probabilities for each sum.
Put only the $3\text{ event outcomes}$ in the denominator.	Use the complete sample-space count: the ticket event is $\frac{3}{8}$ , not $\frac{3}{3}$ or $\frac{1}{3}$ .
Combine $\texttt{HT}$ and $\texttt{TH}$ while treating the remaining patterns equally.	Keep both ordered patterns, or account for their combined probability.
“Not exactly $1$ Head” means $0$ Heads only.	Include every other possible count: $0\text{ or }2$ Heads for $2$ tosses.
Include ticket $6$ in both “at least $6$ ” and its complement.	The complement is less than $6$ ; the $2$ groups cannot share ticket $6$ .
Subtract a decimal from $100$ to find its complement.	Use $1 – 0.375 = 0.625$ , or $100\% – 37.5\% = 62.5\%$ .
Accept a model because its entries total $1$ , even with a negative entry.	Also check that each individual probability lies between $0$ and $1$ .
Assume the complement rule needs independence.	It follows from the full split into $E$ and not $E$ ; independence is not required.
Treat a probability as an exact short-run count.	A model probability gives a chance; the observed number of event hits can vary.

A fast reasonableness check: probabilities must lie between $0$ and $1$ ; an event and its complement must total $1$ ; and the event count cannot exceed the sample-space count.

10 · Try before you reveal

Practice with hints and solutions

Write the sample space or the relevant event outcomes before calculating. Each question has a separate hint and solution, so you can get a small nudge without seeing the full answer.

1. Even on a fair die

A fair $6\text{-sided}$ die is rolled $1\text{ time}$ . Write $S$ and the event $E$ : “an even result.” Find $P(E)$ .

Hint for question 1

List all $6$ numbers, then mark the ones divisible by $2$ .

Solution for question 1

$S = \{1, 2, 3, 4, 5, 6\}$ . $E = \{2, 4, 6\}$ . The outcomes are equally likely, so $P(E) = \frac{3}{6} = \frac{1}{2} = 0.50$ .

2. $\text{Greater than }6$

$1\text{ of the }8$ identical, well-mixed tickets numbered $1\text{-}8$ is drawn. Let $E$ mean “the number is greater than $6$ .” Find $P(E)$ , describe $E^c$ , and find $P(E^c)$ .

Hint for question 2

Does “greater than $6$ ” include ticket $6$ ? Put every number outside $E$ into the complement.

Solution for question 2

$E = \{7, 8\}$ , so $P(E) = \frac{2}{8} = 0.25$ . $E^c = \{1, 2, 3, 4, 5, 6\}$ , meaning “at most $6$ .” Its probability is $1 – 0.25 = 0.75$ .

3. $3\text{ counts do not mean }3\text{ equal chances}$

For $2$ independent fair coin tosses, a student says: “The possible head counts are $0$ , $1$ and $2$ , so exactly $1$ Head has probability $\frac{1}{3}$ .” Correct the reasoning.

Hint for question 3

Use the ordered patterns $\texttt{HH}$ , $\texttt{HT}$ , $\texttt{TH}$ and $\texttt{TT}$ . Which $2$ patterns give $1$ Head?

Solution for question 3

The $4$ ordered patterns are equally likely. Exactly $1$ Head occurs in $\texttt{HT}$ and $\texttt{TH}$ , so the probability is $\frac{2}{4} = \frac{1}{2}$ . The head counts have probabilities $\frac{1}{4}$ , $\frac{1}{2}$ and $\frac{1}{4}$ ; they are not equally likely.

4. A $\text{dice sum of }4$

$2\text{ distinguishable fair dice}$ are rolled independently. List the ordered pairs with sum $4$ . Find $P(\text{sum 4})$ and $P(\text{not sum 4})$ .

Hint for question 4

Find every pair that totals $4$ . There are $36$ equally likely pairs in the whole sample space.

Solution for question 4

The event outcomes are $(1, 3)$ , $(2, 2)$ and $(3, 1)$ . $P(\text{sum 4}) = \frac{3}{36} = \frac{1}{12} \approx 0.0833$ . Its complement contains the other $33$ pairs, so $P(\text{not sum 4}) = 1 – \frac{1}{12} = \frac{11}{12} \approx 0.9167$ .

5. Negate “at most”

A fair die is rolled $1\text{ time}$ . Let $E$ mean “the result is at most $4$ .” Describe $E^c$ and find its probability.

Hint for question 5

“At most $4$ ” includes $4$ . What results remain?

Solution for question 5

$E = \{1, 2, 3, 4\}$ . $E^c = \{5, 6\}$ , or “greater than $4$ .” $P(E^c) = \frac{2}{6} = \frac{1}{3}$ .

6. $\text{Exactly }1\text{ Head in }3\text{ tosses}$

A fair coin is tossed $3\text{ times independently}$ . Use the ordered sample space $\{\texttt{HHH}, \texttt{HHT}, \texttt{HTH}, \texttt{HTT}, \texttt{THH}, \texttt{THT}, \texttt{TTH}, \texttt{TTT}\}$ . Find the event “exactly $1$ Head,” its probability, and its complement.

Hint for question 6

Mark the patterns containing $\text{exactly }1$ $\texttt{H}$ . The complement also includes patterns with $\text{more than }1$ $\texttt{H}$ .

Solution for question 6

$E = \{\texttt{HTT}, \texttt{THT}, \texttt{TTH}\}$ . There are $8$ equally likely patterns, so $P(E) = \frac{3}{8} = 0.375$ . The complement is $\{\texttt{TTT}, \texttt{HHT}, \texttt{HTH}, \texttt{THH}, \texttt{HHH}\}$ : $0,2\text{ or }3$ Heads. $P(E^c) = \frac{5}{8} = 0.625$ .

7. Use the given probabilities

A fictional color model assigns $P(\text{Red})=0.50$ , $P(\text{Blue})=0.30$ and $P(\text{Gold})=0.20$ . What is $P(\text{not Blue})$ ? Explain why $\frac{2}{3}$ is not the answer.

Hint for question 7

Use the complement of Blue. Check whether the $3$ color categories have equal probabilities.

Solution for question 7

$P(\text{not Blue}) = 1 – 0.30 = 0.70$ . The complement consists of Red and Gold, whose probabilities total $0.50 + 0.20 = 0.70$ . The value $\frac{2}{3}$ incorrectly treats all $3$ colors as equally likely.

8. Which probability model is valid?

Each row proposes probabilities for a complete set of $3$ nonoverlapping outcomes. Decide which rows form valid models, and explain.

$3\text{ proposed models}$ for outcomes $1$ , $2$ and $3$ .
Model	Outcome 1	Outcome 2	Outcome 3
A	$0.20$	$0.50$	$0.40$
B	$-0.10$	$0.60$	$0.50$
C	$0.15$	$0.35$	$0.50$

Hint for question 8

Apply both checks: every entry must be from $0$ to $1$ , and the whole row must total $1$ .

Solution for question 8

A is invalid: $0.20 + 0.50 + 0.40 = 1.10$ . B is invalid: its entries total $1$ , but $-0.10$ is a negative probability. C is valid: all entries are within $[0,1]$ and $0.15 + 0.35 + 0.50 = 1$ .

9. Fill the missing probability

A game has $3$ possible outcomes: Win, Draw and Lose. A model gives $P(\text{Win}) = 0.18$ and $P(\text{Draw}) = 0.12$ . Find $P(\text{Lose})$ and $P(\text{not Lose})$ .

Hint for question 9

The $3\text{ probabilities}$ must total $1$ . “Not Lose” includes both Win and Draw.

Solution for question 9

$P(\text{Lose}) = 1 – 0.18 – 0.12 = 0.70$ . $P(\text{not Lose}) = 1 – 0.70 = 0.30$ . Check: $0.18 + 0.12 = 0.30$ .

10. Chance is not a fixed short-run count

A model gives a bonus on $1\text{ play}$ with probability $0.04$ . Find $P(\text{no bonus})$ . Must exactly $4\text{ of the next}$ $100$ independent plays give bonuses under this model?

Hint for question 10

Use $1$ minus the bonus probability. Then distinguish a chance model from a guaranteed observed count.

Solution for question 10

$P(\text{no bonus}) = 1 – 0.04 = 0.96 = 96\%$ . No, the next $100$ plays need not give exactly $4$ bonuses. The number observed can vary; a $4\%$ probability does not force every batch of $100$ to contain $4$ event hits.

11 · Keep the essentials

Quick revision

Define $1\text{ trial}$ Make clear what is performed and what result is recorded.

Write $S$ Include every possible outcome

1\text{ time}

, with enough detail for the question.

Write $E$ Translate the exact condition, including its endpoints.

Check equal likelihoodUse outcome counts only when all outcomes counted in

S

are equally likely.

$\frac{\text{event outcome count}}{\text{whole-space outcome count}}$ The denominator includes outcomes outside

E

Build $E^c$ Include every outcome outside

E

, rather than just one opposite-looking outcome.

Use $1 – P(E)$ Event and complement probabilities always total

1

Explain the chanceUse context and avoid claiming a guaranteed result for the next trial.

Questions students often ask

Must a sample space always have equally likely outcomes?

No. A sample space lists possible outcomes. Equal likelihood is an extra condition needed for the simple $\frac{\text{event outcome count}}{\text{sample-space outcome count}}$ formula. If outcomes have unequal probabilities, use their assigned probabilities.

Does the complement rule require independence?

No. $E$ and $E^c$ describe whether the same event occurs or does not occur. Together they cover the whole sample space without overlap, so their probabilities total $1$ .

Does probability

0.50

guarantee

50

event hits in

100

trials?

No. A finite run can have more or fewer event hits. Under appropriate repeated-trial conditions, relative frequency tends toward the model probability over the long run.

Is “not

\text{exactly }1

” the same as “none”?

No. “Not $\text{exactly }1$ ” includes every possible count $\text{except }1$ . With $2$ coin tosses, it includes $0$ Heads and $2$ Heads. “None” includes only $0$ Heads.

12 · Show that you understand

Final understanding check

A bag $\text{contains }10\text{ identical}$ , well-mixed tokens numbered $1\text{-}10$ , $1\text{ of each}$ . Tokens $1\text{-}5$ are labeled Red, $6\text{-}8$ Blue, and $9\text{-}10$ Gold. $1$ token is drawn without looking, and each numbered token is equally likely.

Write the sample space for the drawn number.
Let $E$ mean “the number is a multiple of $3$ .” List $E$ and calculate $P(E)$ .
List $E^c$ and calculate
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Jitendra Kumar Kumawat

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Raahil Hasan

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Tina Mudarres

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The classes are amazing and my child learnt so much

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alisha gadoya

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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.

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smasher 123

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

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Mosa Al- Samaraie

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AMAZING PRICES GREAT TEACHERS STRAIGHT FORWARD LEARNING STEADY PACE IN TUTORING

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Cael Dagnelie

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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!
NUM8ERS is one of finest tutoring institutes in UAE, Located in Al Barsha 1, Dubai. Close to DUBAI AMERICAN ACADEMY (DAA) & AMERICAN SCHOOL OF DUBAI (ASD).
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