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?"
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?
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 $E$: $P(E') = 1 - P(E)$.
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 $E$
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
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 $P(E')$
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.
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?
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…"
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.
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 $P(E') = 1 - P(E) = 1 - 0.35 = 0.65$.
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. $5/8 = 0.625 = 62.5\%$.
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 .
Worked examples · 3 problems
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.
$S = \{1, 2, 3, 4, 5, 6\}$
$n(S) = 6$
$E = \{2, 4, 6\}$, $n(E) = 3$
$P(\text{even}) = 3/6 = 1/2$
$P(6) = 1/6$
$P(\text{not a 6}) = 1 - 1/6 = 5/6$
A card is drawn from a standard 52-card deck. (a) Find $P(\text{ace})$. (b) Find $P(\text{not an ace})$. (c) Find $P(\text{red card or ace})$.
$n(S) = 52$; 4 aces in the deck
$P(\text{ace}) = 4/52 = 1/13$
$P(\text{not ace}) = 1 - 4/52 = 48/52 = 12/13$
Red cards: 26. Aces: 4. Red aces already counted in both: 2 (hearts ace + diamonds ace).
$n(\text{red or ace}) = 26 + 4 - 2 = 28$
$P = 28/52 = 7/13$
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?
$P(F) = 12/30 = 2/5 = 0.4 = 40\%$
$P(F') = 1 - 2/5 = 3/5 = 0.6 = 60\%$
$P(\text{French}) = 0.4 < P(\text{not French}) = 0.6$
It is more likely the selected student does NOT study French.
Match each event with its probability: (rolling a standard 6-sided die)
Top 3 list: Name THREE checks you should make to ensure a probability answer is correct.
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.
SA 1. A box contains 15 balls: 7 red, 4 white, and 4 black. A ball is selected at random. (a) List the sample space in terms of colours. State $n(S)$. (b) Find $P(\text{red})$ as a fraction. (c) Find $P(\text{not white})$ using the complement. (3 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)
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)
📖 Comprehensive answers (click to reveal)
SA 1 (3 marks): (a) $S = \{\text{red, white, black}\}$; $n(S) = 15$ [1]. (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].
Five timed probability fundamentals questions. Gold tier: 90% + speed.
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