Sec 2 Probability: Sample Spaces & Worked Examples

Secondary 2 G3 Mathematics · Study guide

Build a clear sample space before counting, distinguish outcomes from event labels, and use probabilities to reason about changing groups.

Concept → worked examples → practice14 min read · practice linked below

Probability starts with a clear set of outcomes

A probability measures how likely an event is. It lies between 0 and 1: 0 means impossible and 1 means certain. An outcome is one possible result; an event is a collection of outcomes that meet a condition.

The sample space is the complete set of possible outcomes. When these outcomes are equally likely,

$P(E)=\frac{\text{number of outcomes in }E}{\text{total number of outcomes}}.$

Before you begin: be comfortable simplifying fractions and listing items systematically. Always say what is selected or performed: one ball, one person, one roll or one spin. The sample space must match that action.

“There are three possible colours” does not make each colour’s probability $1/3$ . Equal likelihood is about the underlying outcomes. A bag with many more red balls than blue balls gives different probabilities even though there are only two colour names.

List outcomes without omissions or repetitions

Worked example 1 — a prime number on a fair die

A fair six-sided die is rolled once. Find the probability of obtaining a prime number, and the probability of not obtaining a prime number.

The sample space is $S=\{1,2,3,4,5,6\}$ . Each face is equally likely because the die is fair. The prime-number event is $E=\{2,3,5\}$ ; 1 is not prime.

$P(E)=\frac36=\frac12.$

The complement, “not prime”, consists of the remaining outcomes $\{1,4,6\}$ , so its probability is also $3/6=1/2$ . Equivalently,

$P(\text{not }E)=1-P(E)=1-\frac12=\frac12.$

Check: the event and its complement together contain every outcome exactly once, so their probabilities add to 1. “Not prime” is not the same as “composite”, since it includes 1.

The complement rule works even when the underlying outcomes are not equally likely. An event and its complement still exhaust all possibilities and cannot occur together.

Combine mutually exclusive events carefully

Worked example 2 — one draw from a bag

A bag contains 4 red, 3 blue and 5 green balls. The balls are identical apart from colour and well mixed. One ball is chosen at random. Find the probability that it is red or green.

The equally likely outcomes are the individual balls, not the three colour names. There are $4+3+5=12$ balls in total and $4+5=9$ favourable balls.

$P(\text{red or green})=\frac9{12}=\frac34.$

A single ball cannot be both red and green, so these events are mutually exclusive. We may also add their probabilities: $4/12+5/12=9/12$ .

Check: “red or green” is the complement of “blue”, giving $1-3/12=3/4$ . The denominator stays 12 because only one selection is made from the original bag.

Do not add probabilities blindly when events overlap. On a die, “even” and “greater than 3” overlap at 4 and 6. Their combined event is $\{2,4,5,6\}$ , with probability $4/6$ , not $3/6+3/6$ . Listing the outcomes prevents double-counting.

“Mutually exclusive” means the events cannot occur together in this trial. It does not mean that one event has no influence on another trial; independence is a different idea.

Distinguish a model from results of trials

Worked example 3 — unequal colour regions and experimental probability

A fair spinner has four equal sectors: two red, one blue and one green. In 200 spins it lands on red 92 times. Find the theoretical and experimental probabilities of red, then estimate the number of red results in another 500 spins using the experimental results.

Count the four equally likely sectors. Two of them give red, even though only three colours are present.

The theoretical probability uses the equal-sector model: $P(\text{red})=2/4=1/2$ .
The experimental probability is the relative frequency: $92/200=0.46$ .
Using that experimental estimate, predict $500(0.46)=230$ red results in another 500 spins.

Check: 230 is an estimate, not a guaranteed count. The theoretical model would give an expected count of $500(1/2)=250$ . A difference between observed and theoretical values can arise by chance; these 200 results alone do not prove that the spinner is unfair.

With many repetitions under the same conditions, relative frequencies tend to become more stable. A long run does not force the next spin to “make up” earlier results. The spinner’s construction and the way it is spun still matter.

When the group changes, change the denominator too

Worked example 4 — adding balls to reach a target probability

A bag contains 4 red, 3 blue and 5 green balls. Some red balls are added. After mixing, the probability of selecting a red ball is $1/2$ . How many red balls were added?

Let $x$ be the number added, with $x$ a non-negative integer. The number of red balls becomes $4+x$ . The total becomes $12+x$ , because the new red balls increase both the favourable count and the overall count.

$\frac{4+x}{12+x}=\frac12\quad\Rightarrow\quad2(4+x)=12+x.$

Expand and collect terms: $8+2x=12+x$ , so $x=4$ . Thus 4 red balls were added.

Check: the new bag contains 8 red balls and $8+3+5=16$ balls altogether. The probability is $8/16=1/2$ , as required. The answer is a whole number and non-negative.

There is also a direct count check: a probability of one half requires as many red balls as non-red balls. There are already $3+5=8$ non-red balls, so the red count must rise from 4 to 8.

Explore a bag’s sample space

Change the counts and the event

Each symbol represents one equally likely ball. Choose an event and watch how its favourable count and probability change. Compare “red or green” with the complement of “blue”.

Ball labels identify colours without relying on colour alone. A dark outline marks balls included in the selected event.

Red or green: 9 favourable balls out of 12. Probability = 9/12 = 3/4 = 0.75. Its complement has probability 1/4.

Common mistakes and dependable checks

Assuming event labels are equally likely: count equally likely balls, faces or equal sectors, not just colours or categories.
Using an incomplete sample space: list systematically and check the total. Do not omit zero if it is a possible outcome.
Adding overlapping events: count every favourable outcome once. Direct addition is valid for mutually exclusive events.
Keeping the old total after an addition or removal: update both numerator and denominator as appropriate.
Confusing probability with a count: probability has no unit and must lie from 0 to 1. A predicted number of successes is a separate quantity.
Calling an empty bag probability zero: no ball can be selected, so the experiment is undefined. A valid experiment with no favourable outcomes has probability zero.

Practise counting before calculating

Use the Sec 2 probability practice questions for single selections, combined categories and changing groups. For questions using data tables, also review frequency tables and averages.