What are some real-life applications of probability in decision making?

Probability is widely used in decision making to assess risks and predict outcomes. In real life, it helps in areas such as finance for risk assessment, in weather forecasting to predict weather conditions, and in medicine to determine the likelihood of diseases. Understanding probability allows individuals and organizations to make informed decisions by evaluating the likelihood of various outcomes.

How is probability applied in games of chance?

In games of chance, probability is used to calculate the likelihood of different outcomes. For example, in a game of dice, the probability of rolling a specific number can be calculated by dividing the number of favorable outcomes by the total number of possible outcomes. This helps players understand the odds and make strategic decisions based on the likelihood of certain events occurring.

How does probability theory apply to genetics?

Probability theory is fundamental in genetics, particularly in predicting the inheritance of traits. Mendelian genetics uses probability to determine the likelihood of offspring inheriting certain traits from their parents. For example, Punnett squares are used to calculate the probability of different genotypes and phenotypes in the offspring based on the genetic makeup of the parents.

What role does probability play in quality control processes?

In quality control, probability is used to determine the likelihood of defects in manufacturing processes. Statistical methods such as control charts and sampling plans rely on probability to monitor production processes and ensure product quality. By understanding the probability of defects, companies can implement measures to reduce errors and improve overall quality.

How is probability used in the field of insurance?

Probability is crucial in the insurance industry for assessing risk and setting premiums. Actuaries use probability models to estimate the likelihood of events such as accidents, natural disasters, or health issues. This helps insurance companies determine the appropriate premium rates to charge policyholders, ensuring they can cover potential claims while remaining profitable.

Why is "Adding when independent (or multiplying when mutex)" a common mistake in Probability Applications?

Why it happens: Confusing AND with OR. How to avoid it: OR (mutually exclusive) → ADD. AND (independent) → MULTIPLY. Source: 0580/42 — recurring

Why is "Forgetting to update the denominator on the second draw" a common mistake in Probability Applications?

Why it happens: Doing both draws as if they were the first. How to avoid it: Without replacement: total drops by 1, and the favourable count drops by 1 if the first matched.

Why is "Getting an answer > 1" a common mistake in Probability Applications?

Why it happens: Adding when you should multiply, or counting outcomes wrong. How to avoid it: All probabilities are between 0 and 1. If your answer is outside, recheck.

Why is "Using $\dfrac{\text{favourable}}{\text{total}}$ when outcomes aren't equally likely" a common mistake in Probability Applications?

Why it happens: Defaulting to the simple rule. How to avoid it: The simple rule needs equally-likely outcomes. Otherwise weight each outcome.

Probability Applications – Study Notes & Past Paper Style Questions | Cambridge IGCSE Maths 0580 Extended

Short Notes - Probability Applications

Detailed Study Notes

Detailed notes on Probability for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Probability — Cambridge IGCSE 0580 Maths Extended (2026)

Compute probabilities from equally likely outcomes, complementary events, AND/OR rules, and relative frequency. The base layer underneath tree diagrams and Venn-based probability.

At a glance

Probability $P(A) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$ .
Range: $0 \le P(A) \le 1$ . Express as fraction, decimal, or percentage.
Complement: $P(A') = 1 - P(A)$ .
Mutually exclusive events: $P(A \cup B) = P(A) + P(B)$ .
Independent events: $P(A \cap B) = P(A) \times P(B)$ .
Relative frequency $= \dfrac{\text{number of times event occurs}}{\text{total trials}}$ .
Expected number $= P(A) \times n$ for $n$ trials.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E10.1 — Calculate probability of single events from equally likely outcomes.
E10.2 — Use $P(A') = 1 - P(A)$ .
E10.3 — Calculate probabilities of combined events using addition and multiplication rules.
E10.4 — Use relative frequency as an estimate of probability.

Probability of a single event

$P(A) = \dfrac{\text{favourable outcomes}}{\text{total outcomes}}$ , when all outcomes are equally likely.

Definition. When all outcomes are equally likely: $P(A) = \dfrac{n(A)}{n(\xi)}.$

Range. $0 \le P(A) \le 1$ . $P = 0$ means impossible; $P = 1$ means certain.

Every probability sits between 0 (impossible) and 1 (certain) — writeable as a fraction, decimal or percentage.

Forms. Cambridge accepts fractions, decimals, or percentages. Don't mix notations within one answer.

Worked. Bag with $5$ red, $3$ blue, $2$ green balls. One drawn at random.

$P(\text{red}) = \dfrac{5}{10} = \dfrac{1}{2}$ .
$P(\text{blue or green}) = \dfrac{5}{10} = \dfrac{1}{2}$ .
$P(\text{yellow}) = 0$ (impossible).
$P(\text{red, blue, or green}) = 1$ (certain).

Worked. Two fair dice rolled. Probability of total being $7$ .

Sample space: $36$ ordered pairs.
Favourable: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — $6$ outcomes.
$P = \dfrac{6}{36} = \dfrac{1}{6}$ .

Listing every outcome in a grid makes counting favourable cases straightforward.

Probability = favourable / total (equally likely).
Range $[0, 1]$ .
Use fractions, decimals, or %, not mixed.
List the sample space if it's small.

Complementary events

$P(A') = 1 - P(A)$ . Use when 'NOT $A$ ' is easier to count than $A$ .

Complement rule. $A'$ is the event " $A$ does not happen". $P(A) + P(A') = 1, \quad P(A') = 1 - P(A).$

Use it when. "NOT $A$ " is much easier to count than $A$ itself.

Worked. A bag has $4$ red and $6$ blue balls. Two are drawn without replacement. Probability of "at least one red".

Easier: compute $P(\text{no red}) = P(\text{both blue})$ .
$P(\text{both blue}) = \dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90} = \dfrac{1}{3}$ .
$P(\text{at least one red}) = 1 - \dfrac{1}{3} = \dfrac{2}{3}$ .

The event and its complement fill the whole bar, so P(A′) = 1 − P(A).

Tip. "At least one" almost always begs for the complement: $1 - P(\text{none})$ .

$P(A') = 1 - P(A)$ .
Use for 'at least one' problems.
Sometimes much easier than direct counting.

Combined events: AND, OR

OR (mutually exclusive): add. AND (independent): multiply.

Addition rule (mutually exclusive events). If $A$ and $B$ cannot both happen at once: $P(A \cup B) = P(A) + P(B).$

General addition rule (when they CAN overlap): $P(A \cup B) = P(A) + P(B) - P(A \cap B).$

Multiplication rule (independent events). If $A$ and $B$ are independent (one doesn't affect the other): $P(A \cap B) = P(A) \times P(B).$

Worked (mutually exclusive). Roll a die. $P(\text{prime}) = ?$

Primes on a die: $2, 3, 5$ — three outcomes.
$P = \dfrac{3}{6} = \dfrac{1}{2}$ .

Worked (independent). Roll a die and toss a coin. $P(\text{6 AND heads})$ .

$P(6) = \dfrac{1}{6}$ , $P(\text{H}) = \dfrac{1}{2}$ .
Independent: $P = \dfrac{1}{6} \times \dfrac{1}{2} = \dfrac{1}{12}$ .

Worked (overlapping). Pick a card from a standard deck. $P(\text{red OR ace}).$

$P(\text{red}) = \dfrac{26}{52}$ , $P(\text{ace}) = \dfrac{4}{52}$ , $P(\text{red ace}) = \dfrac{2}{52}$ .
$P = \dfrac{26 + 4 - 2}{52} = \dfrac{28}{52} = \dfrac{7}{13}$ .

Situation	Rule
OR — mutually exclusive	$P(A) + P(B)$
OR — can overlap	$P(A) + P(B) - P(A \cap B)$
AND — independent	$P(A) \times P(B)$

Mutually exclusive OR: add.
Overlapping OR: $P(A) + P(B) - P(A \cap B)$ .
Independent AND: multiply.
Always check if events are independent (or with-replacement).

Relative frequency and expected number

Relative frequency $=$ event count / trials. Expected number $= P \times n$ .

Relative frequency is an experimental estimate of probability when outcomes aren't all equally likely (a biased die, a biased coin): $\text{relative frequency} = \dfrac{\text{number of times event occurred}}{\text{total trials}}.$

As trials $\to \infty$ , relative frequency converges to the true probability (the law of large numbers).

Worked. A spinner is spun $200$ times. Red appears $80$ times.

Estimate of $P(\text{red}) = \dfrac{80}{200} = 0.4$ .

As trials increase, the relative frequency settles toward the true probability (law of large numbers).

Expected number. If an event has probability $P$ and we run $n$ trials, the expected count is: $E = P \times n.$

Worked. A die has $P(6) = \dfrac{1}{6}$ . Roll it $300$ times. Expected number of sixes:

$E = \dfrac{1}{6} \times 300 = 50$ .

Tip. Cambridge often pairs "relative frequency" with "expected number" in a single question — use the relative frequency as the probability estimate, then multiply by $n$ .

Relative frequency = count / total.
Estimates the true probability for biased systems.
Expected number $= P \times n$ .
Law of large numbers: more trials → closer estimate.

Quick recap

$P(A) =$ favourable / total (equally likely).
Range $[0, 1]$ .
Complement: $P(A') = 1 - P(A)$ .
OR (mutually exclusive): add.
AND (independent): multiply.
Relative frequency $\approx$ probability for biased systems.
Expected number $= P \times n$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Probability — measure of how likely an event is, between $0$ and $1$ .
Sample space — set of all possible outcomes.
Mutually exclusive events — events that cannot both occur.
Independent events — events where one does not affect the other.
Relative frequency — experimental estimate of probability.
Expected number — average number of occurrences in $n$ trials: $E = nP$ .

How it’s examined

Probability appears every Paper 2 (3-4 marks) and Paper 4 (6-8 marks). Cambridge stacks single-event probability with combined events and tree diagrams. Examiner reports flag two errors: (i) multiplying probabilities for events that are NOT independent, (ii) confusing relative frequency with theoretical probability.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E10.1-10.4); 0580/22 May/Jun 2024 — Q13 (probability + relative frequency); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Probability Applications

Step-by-step solutions to past-paper-style questions on probability applications, written exactly the way a tutor would explain them at the board.

Question type:

Question patterns to master — Probability Applications

Almost every probability applications exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single probability is asked for — find $P(\dots)$ — from one experiment: a single draw, one combined AND/OR event, a complement, a conditional probability, or a relative frequency.

How to approach it

Pick the matching rule: $\tfrac{\text{favourable}}{\text{total}}$ for one event, multiply for independent AND, add for mutually exclusive OR, $P(A \cup B) = P(A)+P(B)-P(A\cap B)$ when events overlap, and $1 - P(A)$ for 'at least one'.

Common trap

Examiner reports flag adding when events overlap (double-counting the intersection), dividing relative frequency by the number of outcomes instead of total trials, and using the grand total as the denominator for a conditional probability.

Multi-step problem

Recognise it by

Several stages chained together — two draws, a coin choosing a bag, or computing an expected value and then judging fairness — with more than one route to the outcome.

How to approach it

Break the situation into separate paths, find the probability of each path by multiplying along it, then add the mutually exclusive paths. For expected value, compute $\sum x\,P(X=x)$ and only then compare it to the cost.

Common trap

Examiner reports flag computing only one ordering (e.g. $WB$ but not $BW$ ), forgetting to weight each branch, and finding $E(X)$ correctly but not comparing it to the entry cost to judge fairness.

1Probability of a single event
CoreDirect calculation• single
Question
A bag contains $5$ red, $3$ blue, $2$ green balls. Find the probability of drawing a blue ball.
Step-by-step solution
1. Step 1
  $P(\text{event}) = \dfrac{\text{favourable}}{\text{total}}$ .
  $P(\text{blue}) = \dfrac{3}{10}$
Answer
$\dfrac{3}{10}$
2Mutually exclusive events
CoreDirect calculation• Adapted from 0580/22 May/Jun 2024 Q9• mutually exclusive
Question
A spinner has equal sectors numbered $1$ – $8$ . Find $P(\text{multiple of 3 OR multiple of 4})$ .
Step-by-step solution
1. Step 1
  Identify outcomes.
  $\text{Mults of 3: } \{3, 6\};\ \text{Mults of 4: } \{4, 8\}$
2. Step 2
  No overlap → use $P(A) + P(B)$ .
  $P = \dfrac{2}{8} + \dfrac{2}{8} = \dfrac{1}{2}$
Answer
$\dfrac{1}{2}$
3Independent events
ExtendedDirect calculation• independent
Question
A coin is flipped and a fair die is rolled. Find $P(\text{tails AND a 6})$ .
Step-by-step solution
1. Step 1
  Independent events: multiply the individual probabilities.
  $P = \dfrac{1}{2} \times \dfrac{1}{6} = \dfrac{1}{12}$
Answer
$\dfrac{1}{12}$
4Without replacement
ExtendedDirect calculation• dependent
Question
A bag has $4$ red and $6$ blue balls. Two are drawn without replacement. Find $P(\text{both red})$ .
Step-by-step solution
1. Step 1
  First draw: $\dfrac{4}{10}$ . After removing one red, $3$ red and $9$ total remain.
  $P = \dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90} = \dfrac{2}{15}$
Answer
$\dfrac{2}{15}$
5Complementary probability
ExtendedDirect calculation• complement
Question
A factory produces faulty bulbs with probability $0.02$ . Find $P(\text{at least one fault in } 5 \text{ bulbs})$ .
Step-by-step solution
1. Step 1
  Use the complement: $P(\text{none}) = 0.98^{5}$ .
  $0.98^{5} \approx 0.9039$
2. Step 2
  Subtract from 1.
  $P(\text{at least one}) \approx 1 - 0.9039 = 0.0961$
Answer
$\approx 0.0961$
6Estimating probability from a frequency table
CoreDirect calculation• Adapted from 0580/22 Oct/Nov 2023 Q14• relative frequency, experimental
Question
A four-sided spinner was spun $200$ times. Results: $1$ → $42$ , $2$ → $58$ , $3$ → $46$ , $4$ → $54$ . Use the results to estimate the probability that the spinner lands on $2$ .
Step-by-step solution
1. Step 1
  Relative frequency = $\dfrac{\text{number of times event occurred}}{\text{total trials}}$ .
2. Step 2
  Substitute the values.
  $P(2) \approx \dfrac{58}{200} = \dfrac{29}{100} = 0.29$
Answer
$0.29$ (or $\dfrac{29}{100}$ )
Examiner tip
The examiner report flags that candidates often divide by $4$ (the number of outcomes) instead of $200$ (the total trials). Always read the question — empirical probability uses the total number of trials as the denominator.
7Using $P(A) + P(A') = 1$
CoreDirect calculation• complement
Question
The probability that it rains on a given day in October is $0.35$ . Find the probability that it does NOT rain on that day.
Step-by-step solution
1. Step 1
  Use the complement rule: $P(A') = 1 - P(A)$ .
  $P(\text{no rain}) = 1 - 0.35 = 0.65$
Answer
$0.65$
8Conditional probability from a table
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2023 Q9• conditional
Question
A survey of $80$ students asked whether they like Maths (M) and Science (S). $48$ like Maths, $36$ like Science, $20$ like both. A student who likes Science is chosen at random. Find the probability that this student also likes Maths.
Step-by-step solution
1. Step 1
  Restrict to the conditioning set: only the $36$ Science-lovers matter.
2. Step 2
  Of those $36$ , how many also like Maths? That is $n(M \cap S) = 20$ .
  $P(M \mid S) = \dfrac{n(M \cap S)}{n(S)} = \dfrac{20}{36}$
3. Step 3
  Simplify.
  $= \dfrac{5}{9}$
Answer
$\dfrac{5}{9}$
Examiner tip
The examiner report consistently notes that the denominator must be $n(S)$ (the conditioning event), not $80$ (the total). Writing $\dfrac{20}{80}$ scores zero.
9Without replacement — exactly one of each colour
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q11• without replacement, two-draw
Question
A box contains $7$ white and $5$ black counters. Two counters are taken without replacement. Find the probability that one is white and the other is black.
Step-by-step solution
1. Step 1
  Two valid orderings give 'one of each': $WB$ or $BW$ .
2. Step 2
  Path $WB$ : first $W$ then $B$ .
  $P(WB) = \dfrac{7}{12} \times \dfrac{5}{11} = \dfrac{35}{132}$
3. Step 3
  Path $BW$ : first $B$ then $W$ .
  $P(BW) = \dfrac{5}{12} \times \dfrac{7}{11} = \dfrac{35}{132}$
4. Step 4
  Add the two mutually exclusive paths.
  $P(\text{one of each}) = \dfrac{35}{132} + \dfrac{35}{132} = \dfrac{70}{132} = \dfrac{35}{66}$
Answer
$\dfrac{35}{66}$
Examiner tip
The examiner report flags that candidates often compute only one path (e.g. $WB$ ) and forget the symmetric path $BW$ . Both orderings must be added.
10Expected value from a frequency distribution
ExtendedMulti-step problem• expected value
Question
A game costs $\$ 2 $to play. The probabilities of winning$ $0 $,$ $1 $,$ $3 $and$ $10 $are$ 0.5,\ 0.3,\ 0.15 $and$ 0.05$ respectively. Find the expected winnings per game and state whether the game is fair.
Step-by-step solution
1. Step 1
  Expected value $E(X) = \sum x \cdot P(X = x)$ .
  $E(X) = 0(0.5) + 1(0.3) + 3(0.15) + 10(0.05)$
2. Step 2
  Compute each term.
  $E(X) = 0 + 0.3 + 0.45 + 0.5 = 1.25$
3. Step 3
  Expected winnings $= \$ 1.25 $. Since the cost to play is$ $2 $, the player loses$ $0.75$ per game on average — the game is not fair.
Answer
Expected winnings $= \$ 1.25 $per game; not fair (loss of$ $0.75$ on average).
Examiner tip
A fair game has expected winnings equal to the cost to play. Many candidates compute $E(X)$ correctly but forget to compare it to the entry cost.
11OR-rule when events overlap
ExtendedDirect calculation• union, inclusion-exclusion
Question
A card is drawn from a standard 52-card deck. Find the probability that the card is a $\heartsuit$ OR a king.
Step-by-step solution
1. Step 1
  These events overlap — the King of Hearts is both. Use $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .
2. Step 2
  Substitute.
  $P(\heartsuit \cup K) = \dfrac{13}{52} + \dfrac{4}{52} - \dfrac{1}{52} = \dfrac{16}{52}$
3. Step 3
  Simplify.
  $= \dfrac{4}{13}$
Answer
$\dfrac{4}{13}$
Examiner tip
The examiner report flags that candidates routinely apply $P(A) + P(B)$ here and arrive at $\dfrac{17}{52}$ , double-counting the King of Hearts. The events are NOT mutually exclusive — always subtract the intersection.
12Combined independent and dependent events
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2023 Q10• combined, challenge
Question
A coin is flipped: if it lands heads, a ball is drawn from Bag A (containing $3$ red, $2$ blue); if tails, from Bag B (containing $1$ red, $4$ blue). Find the probability that a red ball is drawn.
Step-by-step solution
1. Step 1
  There are two ways to draw red: via Bag A or via Bag B. These are mutually exclusive.
2. Step 2
  Path H then red from A.
  $P(H \cap R) = \dfrac{1}{2} \times \dfrac{3}{5} = \dfrac{3}{10}$
3. Step 3
  Path T then red from B.
  $P(T \cap R) = \dfrac{1}{2} \times \dfrac{1}{5} = \dfrac{1}{10}$
4. Step 4
  Add the two paths (total probability theorem).
  $P(R) = \dfrac{3}{10} + \dfrac{1}{10} = \dfrac{4}{10} = \dfrac{2}{5}$
Answer
$\dfrac{2}{5}$
Examiner tip
This is a common 6-mark question. The examiner report flags that some candidates compute $\dfrac{4}{10}$ but forget to simplify; others forget to weight each bag by $\dfrac{1}{2}$ and instead add the conditional probabilities directly.

Key Formulae — Probability Applications

The formulae you need to memorise for probability applications on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Probability of an event
$P(A) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
When to use
Equally likely outcomes.
Mutually exclusive (OR)
$P(A \cup B) = P(A) + P(B)$
When to use
When $A$ and $B$ cannot both happen.
Independent (AND)
$P(A \cap B) = P(A) \times P(B)$
When to use
When $A$ does not affect the probability of $B$ .
Complement
$P(A') = 1 - P(A)$
When to use
When 'NOT happening' or 'at least one' is easier than direct.

Key Definitions and Keywords — Probability Applications

Definitions to memorise and the exact keywords mark schemes credit for probability applications answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Event
Examiner keyword
A set of outcomes from a probability experiment.
Independent events
Examiner keyword
Events where the outcome of one does not affect the other.
Mutually exclusive events
Examiner keyword
Events that cannot both occur at the same time. $P(A \cap B) = 0$ .
With/without replacement
Examiner keyword
With replacement: the item goes back, probabilities unchanged. Without: the item is kept, probabilities change for the next draw.
Relative frequency
An estimate of probability based on observed outcomes: $\dfrac{\text{observed}}{\text{trials}}$ .

Common Mistakes and Misconceptions — Probability Applications

The traps other students keep falling into on probability applications questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Adding when independent (or multiplying when mutex)
0580/42 — recurring
Why it happens
Confusing AND with OR.
How to avoid it
OR (mutually exclusive) → ADD. AND (independent) → MULTIPLY.
✕Forgetting to update the denominator on the second draw
Why it happens
Doing both draws as if they were the first.
How to avoid it
Without replacement: total drops by 1, and the favourable count drops by 1 if the first matched.
✕Getting an answer > 1
Why it happens
Adding when you should multiply, or counting outcomes wrong.
How to avoid it
All probabilities are between $0$ and $1$ . If your answer is outside, recheck.
✕Using $\dfrac{\text{favourable}}{\text{total}}$ when outcomes aren't equally likely
Why it happens
Defaulting to the simple rule.
How to avoid it
The simple rule needs equally-likely outcomes. Otherwise weight each outcome.

Probability Applications — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Probability Applications

Short Notes - Probability Applications

Detailed Study Notes

At a glance

What you’ll learn

Probability of a single event

Complementary events

Combined events: AND, OR

Relative frequency and expected number

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Probability Applications

1Probability of a single event

2Mutually exclusive events

3Independent events

4Without replacement

5Complementary probability

6Estimating probability from a frequency table

7Using P(A)+P(A′)=1P(A) + P(A') = 1P(A)+P(A′)=1

8Conditional probability from a table

9Without replacement — exactly one of each colour

10Expected value from a frequency distribution

11OR-rule when events overlap

12Combined independent and dependent events

Probability of an event

Mutually exclusive (OR)

Independent (AND)

Complement

Event

Independent events

Mutually exclusive events

With/without replacement

Relative frequency

✕Adding when independent (or multiplying when mutex)

✕Forgetting to update the denominator on the second draw

✕Getting an answer > 1

✕Using favourabletotal\dfrac{\text{favourable}}{\text{total}}totalfavourable​ when outcomes aren't equally likely

Probability Applications — frequently asked questions

Probability Applications

Short Notes - Probability Applications

Detailed Study Notes

At a glance

What you’ll learn

Probability of a single event

Complementary events

Combined events: AND, OR

Relative frequency and expected number

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

Question patterns to master — Probability Applications

1Probability of a single event

2Mutually exclusive events

3Independent events

4Without replacement

5Complementary probability

6Estimating probability from a frequency table

7Using P(A)+P(A′)=1P(A) + P(A') = 1P(A)+P(A′)=1

8Conditional probability from a table

9Without replacement — exactly one of each colour

10Expected value from a frequency distribution

11OR-rule when events overlap

12Combined independent and dependent events

Probability of an event

Mutually exclusive (OR)

Independent (AND)

Complement

Event

Independent events

Mutually exclusive events

With/without replacement

Relative frequency

✕Adding when independent (or multiplying when mutex)

✕Forgetting to update the denominator on the second draw

✕Getting an answer > 1

✕Using favourabletotal\dfrac{\text{favourable}}{\text{total}}totalfavourable​ when outcomes aren't equally likely

Probability Applications — frequently asked questions

7Using $P(A) + P(A') = 1$

✕Using $\dfrac{\text{favourable}}{\text{total}}$ when outcomes aren't equally likely

7Using $P(A) + P(A') = 1$

✕Using $\dfrac{\text{favourable}}{\text{total}}$ when outcomes aren't equally likely