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Relative Frequency & Probability · L1 of 4 ~45 min MST-12-S2-09 ⚡ +85 XP available

Probability Fundamentals

Probability puts a number between 0 and 1 on the likelihood of any event. To find it, you need to count the favourable outcomes and the total possible outcomes, and the complement rule lets you sidestep counting by asking "what's the probability it doesn't happen?"

Today's hook, A bag contains 3 red, 5 blue and 2 yellow marbles. You draw one without looking. What is the probability of NOT drawing blue? Think about it, there's a shortcut that avoids counting non-blue marbles directly.

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Orient and prepare

Capture your first idea, review the formulas and preview the lesson language.

Recall, your gut answer first

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A bag contains 3 red, 5 blue and 2 yellow marbles. You draw one marble at random. Write your gut answers to these, no calculating yet:

Is picking a blue marble more or less likely than not picking blue?
How would you calculate P(not blue), what's the shortcut?
What is the probability of picking a green marble?

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The key formulas you need to own

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Probability is always a number between 0 and 1. $P(E) = \frac{n(E)}{n(S)}$ when all outcomes are equally likely. The complement rule $P(E') = 1 - P(E)$ is the fastest route when it's easier to count what isn't in the event.

$n(E)$ = number of outcomes in event $E$. $n(S)$ = total outcomes in the sample space $S$. Together: $P(E) = n(E)/n(S)$. The complement $E'$ is everything in $S$ that is NOT in $E$: $P(E') = 1 - P(E)$.

$0 \leq P(E) \leq 1$ | $P(E) + P(E') = 1$ | $P(\text{impossible}) = 0$ | $P(\text{certain}) = 1$

Three equivalent forms

$P(E) = 1/4 = 0.25 = 25\%$. All three are correct. HSC answers may require any form, check what format the question asks for.

Use complement when it's faster

If the event has many outcomes, count its complement (fewer outcomes) and subtract from 1. "Not blue" in a bag of 3+5+2 = 10 marbles: P(not blue) = 1 − 5/10 = 5/10 = 1/2.

Equally likely must be assumed

The formula $P(E) = n(E)/n(S)$ only works when every outcome in $S$ is equally likely (e.g., fair coin, well-shuffled deck, unbiased die). Always check this assumption.

What you'll master

Know

Key facts

$0 \leq P(E) \leq 1$; impossible = 0; certain = 1
Sample space $S$ = set of all possible outcomes; $n(S)$ = number of outcomes
$P(E) = n(E)/n(S)$ for equally likely outcomes
Complement: $P(E') = 1 - P(E)$, where $E'$ = all outcomes NOT in $E$

Understand

Concepts

Why probability is always between 0 and 1
What "equally likely outcomes" means and when to check it
Why $P(E) + P(E') = 1$ must always be true
When the complement rule is more efficient than direct counting

Can do

Skills

List a sample space for a simple experiment
Count $n(S)$ and $n(E)$ to calculate $P(E)$
Express probabilities as fractions, decimals and percentages
Apply the complement rule to find $P(E')$

Key terms

Probability $P(E)$A number from 0 to 1 measuring how likely event $E$ is to occur. 0 = impossible; 1 = certain.

Sample space $S$The complete set of all possible outcomes of a chance experiment. E.g. rolling a die: $S = \{1, 2, 3, 4, 5, 6\}$.

Event $E$A subset of the sample space, one or more outcomes we're interested in. E.g. rolling an even number: $E = \{2, 4, 6\}$.

Equally likely outcomesWhen every outcome in $S$ has the same probability of occurring. Required for the formula $P(E) = n(E)/n(S)$ to be valid.

Complement $E'$The event that $E$ does NOT occur, all outcomes in $S$ that are not in $E$. $P(E') = 1 - P(E)$.

$n(E)$ and $n(S)$$n(E)$ = number of outcomes in event $E$. $n(S)$ = total outcomes in the sample space.

Sample Spaces and Listing Outcomes

Build the first core idea, then test it with a quick check.

Sample Spaces and Listing Outcomes

core concept

Before you can calculate any probability, you must be able to list or count the full set of possible outcomes, the sample space. A systematic approach prevents you from missing outcomes.

Strategies for listing a sample space:

List all outcomes: For small experiments, write every possibility: rolling a die $S = \{1, 2, 3, 4, 5, 6\}$, $n(S) = 6$.
Systematic listing: For multi-step experiments, list all combinations in order. Flipping two coins: HH, HT, TH, TT, $n(S) = 4$.
Counting without listing: For larger experiments, use the multiplication principle: a die AND a coin = $6 \times 2 = 12$ outcomes.

Determining $n(S)$ from a description: Read carefully. "A card is drawn from a standard 52-card deck" → $n(S) = 52$. "Two dice are rolled" → $n(S) = 6 \times 6 = 36$. "A letter is chosen from the word PROBABILITY" → count the letters: P-R-O-B-A-B-I-L-I-T-Y = 11 letters (not 11 unique, but 11 total, each position is one outcome). If selecting letters without replacement, context matters.

Events are subsets: Once you have $S$, an event $E$ is any subset. "Rolling a number greater than 4" from $S = \{1,2,3,4,5,6\}$: $E = \{5, 6\}$, $n(E) = 2$. $P(E) = 2/6 = 1/3$.

Sample space: the set of all possible outcomes of an experiment. List outcomes systematically using ordered lists or tables. P(event) = favourable outcomes ÷ total outcomes. All probabilities are between 0 and 1 inclusive.

Pause, copy the sample space definition (complete set of all possible outcomes), the probability formula P(A) = favourable outcomes ÷ total outcomes, and the range rule: all probabilities lie between 0 (impossible) and 1 (certain) inclusive into your book.

Quick check: A spinner has 8 equal sections numbered 1–8. What is the probability of landing on a prime number?

Calculating Probability, the Formula and the Complement

Connect the next method to the foundation and check your reasoning.

Calculating Probability, the Formula and the Complement

core concept

The sample space is the complete list of all possible outcomes, and each outcome is equally likely when working with theoretical probability. To find the probability of an event, count the favourable outcomes and divide by the total: P(A) = favourable ÷ total. The complement rule P(not A) = 1 − P(A) is useful when it is easier to count the outcomes that are NOT in the event.

The probability formula $P(E) = n(E)/n(S)$ and the complement rule $P(E') = 1 - P(E)$ are the two most-used tools in this topic. Knowing when to use each one is the key skill.

Using the formula directly:

1. Identify $n(S)$ (total outcomes) and $n(E)$ (favourable outcomes). 2. Divide: $P(E) = n(E)/n(S)$. 3. Simplify and convert to required form (fraction/decimal/%).

Using the complement:

1. Identify $P(E)$ (probability of the event). 2. Apply $P(E') = 1 - P(E)$. Use this when $n(E')$ is harder to count directly, or when the question asks "probability of NOT…"

Marble example: Bag: 3 red, 5 blue, 2 yellow. $n(S) = 10$.
Direct: $P(\text{blue}) = 5/10 = 1/2$.
Complement: $P(\text{not blue}) = 1 - 5/10 = 5/10 = 1/2$.
Or: count non-blue directly: $n(\text{not blue}) = 3 + 2 = 5$; $P = 5/10 = 1/2$. Both give the same answer, use whichever is faster.

Common error, fractions vs percentages: $P(E) = 0.3$ is the same as $P(E) = 30\% = 3/10$. Never write $P(E) = 30$ (not a fraction/decimal). Probabilities greater than 1 or less than 0 are always wrong.

Probability formula: P(A) = number of favourable outcomes ÷ total outcomes. Complement: P(not A) = 1 − P(A). Use the complement when it is easier to count outcomes that are NOT in the event.

Pause, copy P(A) = favourable ÷ total outcomes, the complement rule P(not A) = 1 − P(A), and note when to use the complement: when counting non-A outcomes is easier than counting A outcomes into your book.

True or false: If $P(\text{rain tomorrow}) = 0.35$, then the probability it does NOT rain tomorrow is $0.65$, because $P(E') = 1 - P(E) = 1 - 0.35 = 0.65$.

Converting Between Fractions, Decimals and Percentages

Apply the third core idea and confirm the key calculation.

Converting Between Fractions, Decimals and Percentages

core concept

P(A) = favourable ÷ total and P(not A) = 1 − P(A) are the two foundational rules. In an HSC answer, probabilities may be expressed as fractions (exact), decimals (easier for calculation), or percentages (for communication), all three are accepted, and questions may give data in any form. Convert fluently: fraction → decimal (divide), decimal → percentage (multiply by 100).

NESA questions may ask you to express a probability as a fraction, decimal or percentage. All three are equivalent, the skill is converting between them accurately and quickly.

Converting probabilities:

Fraction → Decimal: Divide numerator by denominator. $3/8 = 0.375$.
Decimal → Percentage: Multiply by 100. $0.375 \to 37.5\%$.
Percentage → Fraction: Divide by 100 and simplify. $75\% = 75/100 = 3/4$.
Fraction → Percentage: Divide to get decimal, then multiply by 100. $5/8 = 0.625 = 62.5\%$.

Which form to use? HSC questions usually specify. If not: fractions are exact and preferred for simple ratios; decimals are practical for calculations; percentages are natural in everyday contexts. All three earn full marks if correct.

Rounding: Leave probability answers as exact fractions unless the question asks for a decimal or percentage, in which case, follow any rounding instruction given. Unrounded fractions are always safest for intermediate steps.

Probabilities can be expressed as fractions (exact), decimals (for calculation), or percentages (for communication). To convert: fraction → decimal (divide), decimal → percentage (× 100). All three forms are accepted in HSC answers.

Pause, copy the three probability forms and conversions: fraction (exact; divide numerator by denominator to convert), decimal (multiply by 100 to get percentage), percentage (divide by 100 to get decimal), and note all three are accepted in HSC answers into your book.

Fill the blanks: A probability of $\frac{3}{5}$ is equal to as a decimal and %. Its complement is .

Work through the examples

Follow the worked solutions, then transfer the method to the final quick tasks.

Worked examples · 3 problems

PROBLEM 1 · LISTING OUTCOMES AND FINDING PROBABILITY

A standard die is rolled. (a) List the sample space $S$. (b) Find $P(\text{even number})$. (c) Find $P(\text{not a 6})$ using the complement.

Part (a), sample space
$S = \{1, 2, 3, 4, 5, 6\}$
$n(S) = 6$

All six outcomes are equally likely on a fair die.

PROBLEM 2 · CARDS, USING COMPLEMENT

A card is drawn from a standard 52-card deck. (a) Find $P(\text{ace})$. (b) Find $P(\text{not an ace})$. (c) Find $P(\text{red card or ace})$.

Part (a), P(ace)
$n(S) = 52$; 4 aces in the deck
$P(\text{ace}) = 4/52 = 1/13$

A standard deck has 52 cards: 4 suits × 13 values. There is one ace per suit.

PROBLEM 3 · CONVERTING AND COMPLEMENTING

In a class of 30 students, 12 study French and the rest don't. (a) What is $P(\text{studies French})$ as a fraction, decimal and percentage? (b) What is $P(\text{does not study French})$? (c) A student is selected at random. Is it more likely they study French or that they don't?

Part (a), P(French)
$P(F) = 12/30 = 2/5 = 0.4 = 40\%$

Simplify $12/30$ by dividing by 6: $2/5$. Then convert: $2 \div 5 = 0.4$; $0.4 \times 100 = 40\%$.

Match each event with its probability: (rolling a standard 6-sided die)

Rolling a number greater than 2
Rolling a 7
Rolling a number from 1 to 6
Rolling an odd number

P = 1 (certain)
P = 3/6 = 1/2
P = 4/6 = 2/3
P = 0 (impossible)

Top 3 list: Name THREE checks you should make to ensure a probability answer is correct.

Show what you can do

Answer the exam-style questions, then compare your working with the model answers.

Short answer, exam-style questions

show all working

ApplyBand 33 marks

SA 1. A box contains 15 balls: 7 red, 4 white, and 4 black. A ball is selected at random. (a) State the colour categories, and state how many equally likely outcomes there are, $n(S)$. (b) Find $P(\text{red})$ as a fraction. (c) Find $P(\text{not white})$ using the complement. (3 marks)

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ApplyBand 33 marks

SA 2. A letter is chosen at random from the word STATISTICS. (a) How many letters are there in total? (b) Find $P(\text{choosing the letter T})$. (c) Express this probability as a decimal and as a percentage. (3 marks)

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AnalyseBand 44 marks

SA 3. In a survey of 200 people, 85 preferred tea, 70 preferred coffee, and 45 preferred neither. A person is selected at random. (a) Find $P(\text{prefers tea or coffee})$. (b) Find $P(\text{prefers neither})$ and verify using the complement. (c) Is it more likely that a person prefers tea or coffee than neither? Justify with calculations. (4 marks)

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📖 Comprehensive answers (click to reveal)

SA 1 (3 marks): (a) The three colour categories are red, white and black. The equally likely outcomes are the 15 individual balls, so $n(S) = 15$ [1]. (The colours are not themselves equally likely, which is why the outcomes counted are the balls, not the three labels.) (b) $P(\text{red}) = 7/15$ [1]. (c) $P(\text{not white}) = 1 - 4/15 = 11/15$ [1]. (Alternatively: $n(\text{not white})=11$, $P=11/15$.)

SA 2 (3 marks): (a) STATISTICS has 10 letters [1]. (b) T appears 3 times: $P(T) = 3/10$ [1]. (c) $3/10 = 0.3 = 30\%$ [1].

SA 3 (4 marks): (a) $n(\text{tea or coffee})=85+70=155$; $P=155/200=31/40=0.775$ [1]. (b) $P(\text{neither})=45/200=9/40=0.225$ [1]. Verify: $1-155/200=45/200$ ✓ [1]. (c) $0.775 > 0.225$, so yes, it is more than 3 times more likely that a person prefers tea or coffee than neither [1].

Retrieve, reflect and finish

Revisit the lesson idea, use the module challenge and save your completion.

Revisit your thinking

Marble bag (3 red, 5 blue, 2 yellow), $n(S) = 10$. P(not blue) = $1 - 5/10 = 5/10 = 1/2$. The "shortcut" is the complement rule, count the complement (5 non-blue marbles) or subtract from 1. P(green) = 0/10 = 0, impossible, because there are no green marbles in the bag.

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Relative Frequency & Probability · Maths Standard