Probability becomes difficult when the event structure is hidden. The arithmetic itself is often short. The harder work is deciding whether events can happen together, whether one event changes another, which branches belong to the same outcome and whether the question asks for “and”, “or”, “at least one” or “not”.
This fifteenth Secondary 4 Mathematics Learning Guide develops probability as a system for representing uncertainty before calculating it. It belongs to the Secondary Mathematics Hub and S1–S4 Capability Map. It covers single events, complements, combined events, possibility diagrams, tree diagrams, mutually exclusive events and independent events.
Current syllabus connection: the 2026 O-Level Mathematics 4052 and 2027 SEC G3 Mathematics K310 syllabus structures include probability of single events, simple combined events, possibility diagrams, tree diagrams, and addition and multiplication rules for mutually exclusive and independent events. The examples below are original teaching problems rather than reproduced examination questions.
Probability is a number between 0 and 1
A probability of 0 describes an impossible event. A probability of 1 describes a certain event. Values between them describe varying degrees of chance. In percentage form, 0.25 is 25% and 0.8 is 80%.
A calculated probability outside the interval from 0 to 1 signals a structural or arithmetic error. This is one of the fastest verification checks in the topic.
Equally likely outcomes: favourable over total
When all outcomes in a simple sample space are equally likely, probability can be calculated as:
P(event) = number of favourable outcomes / total number of possible outcomes.
The phrase “equally likely” matters. If outcomes have different chances, counting labels alone does not give the probability.
Worked Example 1 | One event from a simple sample space
A fair six-sided die is rolled. Find the probability of obtaining a number greater than 4.
The sample space is {1,2,3,4,5,6}. The favourable outcomes are {5,6}. Therefore:
P(number > 4) = 2/6 = 1/3.
The answer is sensible because it lies between 0 and 1 and there are two favourable outcomes among six equally likely outcomes.
The complement rule: sometimes “not” is the shorter route
An event and its complement cover the whole sample space without overlap. Therefore:
P(A′) = 1 − P(A).
This is especially useful for “at least one” questions, where the complement may be “none”. Instead of adding many successful cases, calculate the single failure case and subtract from 1.
Worked Example 2 | At least one success
A fair coin is tossed three times. Find the probability of obtaining at least one head.
The complement is obtaining no heads, which means TTT. Its probability is (1/2)³ = 1/8.
P(at least one head) = 1 − 1/8 = 7/8.
This is shorter and less error-prone than separately listing exactly one, exactly two and exactly three heads.
“And” often follows one path
In a tree diagram, an outcome involving one event followed by another lies along one complete path. The probability of that path is found by multiplying branch probabilities.
For independent events, the probability of A and B can be written as P(A)P(B). The multiplication reflects the need for both stages to occur along that route.
Independent events: one outcome does not change the next probability
If a fair coin is tossed twice, the second toss remains fair regardless of the first result. The branch probabilities repeat. This is independence.
Independence should not be assumed merely because two stages happen one after another. If an item is removed from a bag and not replaced, the composition of the bag changes, so later probabilities usually change.
Worked Example 3 | With replacement
A bag contains 3 red and 2 blue counters. One counter is chosen at random, replaced, and then a second counter is chosen. Find the probability of two red counters.
Because the first counter is replaced, the bag returns to its original composition before the second draw. Each red probability is 3/5.
P(RR) = 3/5 × 3/5 = 9/25.
Worked Example 4 | Without replacement
Use the same bag, but do not replace the first counter. Find the probability of two red counters.
The first red probability is 3/5. After one red is removed, 2 red and 2 blue counters remain, so the second red probability is 2/4.
P(RR) = 3/5 × 2/4 = 3/10.
The changed denominator records the changed sample space. A tree diagram makes this dependency visible.
“Or” usually means combine separate successful routes
If two mutually exclusive outcomes cannot occur together, the probability of one or the other is found by adding their probabilities.
On a tree diagram, calculate the probability of each successful complete path, then add those path probabilities. Multiply along a path; add across alternative paths.
Along a path: multiply. Across alternative successful paths: add.
Worked Example 5 | Exactly one red
From the 3-red, 2-blue bag, two counters are selected without replacement. Find the probability of exactly one red.
There are two successful routes: red then blue, or blue then red.
P(RB) = 3/5 × 2/4 = 3/10.
P(BR) = 2/5 × 3/4 = 3/10.
P(exactly one red) = 3/10 + 3/10 = 3/5.
The routes are mutually exclusive complete outcomes: the same pair of ordered draws cannot be both RB and BR.
Mutually exclusive is not the same as independent
Mutually exclusive events cannot both occur in the same trial. For one die roll, “roll a 2” and “roll a 5” are mutually exclusive.
Independent events are events where knowing one occurred does not change the probability of the other. Two tosses of a fair coin are a standard example.
These ideas should not be merged. If two events with positive probability are mutually exclusive, occurrence of one makes the other impossible, so they are not independent.
Possibility diagrams: build the full sample space
A possibility diagram is useful when two simple outcomes combine, such as rolling two dice or choosing one option from each of two lists. Each cell represents one ordered combination.
With two fair six-sided dice there are 6 × 6 = 36 equally likely ordered pairs. A sum of 7 occurs in six cells: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Therefore P(sum 7) = 6/36 = 1/6.
Do not treat possible sums 2 through 12 as eleven equally likely outcomes. They have different numbers of supporting ordered pairs.
Worked Example 6 | Sum greater than 9 on two dice
The successful sums are 10, 11 and 12. They occur in 3, 2 and 1 ordered pairs respectively, giving 6 successful outcomes out of 36.
P(sum > 9) = 6/36 = 1/6.
The same answer as the sum-of-7 example does not mean the events are identical. They simply occupy the same number of equally likely cells.
Tree diagrams: branch probabilities leaving a node sum to 1
At each stage of a complete tree, the next possible branches should cover all outcomes from that node. Their probabilities therefore sum to 1.
If a first branch is labelled success with probability 0.7, and the only alternative is failure, the failure probability is 0.3. This local sum-to-one check catches many missing or mistyped branches.
Worked Example 7 | A two-stage quality-control model
A component passes Test 1 with probability 0.8. If it passes Test 1, it passes Test 2 with probability 0.9. If it fails Test 1, a retest route passes Test 2 with probability 0.4. Find the probability that the component passes Test 2.
There are two successful routes:
P(pass Test 1 and pass Test 2) = 0.8 × 0.9 = 0.72.
P(fail Test 1 and then pass Test 2) = 0.2 × 0.4 = 0.08.
P(pass Test 2) = 0.72 + 0.08 = 0.80.
The second-stage probabilities differ depending on the first branch, so this is not a situation where one repeated independent probability should be used automatically.
Experimental probability: evidence from observed trials
If an event occurs 37 times in 100 trials, its experimental probability is 37/100 = 0.37. This is an observed proportion, not proof that the theoretical probability is exactly 0.37.
With more trials under stable conditions, experimental proportions may provide a more stable estimate of the underlying chance. But a probability describes uncertainty; it does not guarantee the outcome of the next trial.
Expected frequency: turn a probability into a long-run count
If an event has probability 0.3 and the experiment is repeated 500 times under comparable conditions, the expected frequency is 0.3 × 500 = 150. The actual observed count may differ.
The expectation is a model-based long-run quantity. It should not be described as a certainty that the event will occur exactly 150 times.
Worked Example 8 | Reverse probability from expected frequency
An event is expected to occur 84 times in 240 trials. Find the probability used in the model.
p = 84/240 = 0.35.
Check: 0.35 × 240 = 84.
Probability language map
| Phrase | Useful representation |
|---|---|
| and | One complete path or intersection |
| or | Alternative successful outcomes; check overlap |
| at least one | Often 1 − P(none) |
| exactly one | Add routes with one success and all other required failures |
| without replacement | Update later branch probabilities |
| with replacement | Original composition is restored |
| mutually exclusive | Cannot occur together in the same trial |
| independent | One occurrence does not alter the other’s probability |
Common failure modes
| Error | Cause | Repair |
|---|---|---|
| Adds probabilities along one tree path | “And” structure missed | Multiply branch probabilities on the same path |
| Multiplies alternative outcomes together | “Or” structure missed | Add mutually exclusive complete routes |
| Repeats branch probability without replacement | Sample space not updated | Rewrite the remaining composition after each draw |
| Treats mutually exclusive as independent | Definitions merged | Ask whether events can occur together and whether one changes the other |
| Lists sums rather than equally likely dice pairs | Sample space compressed incorrectly | Use a possibility diagram |
| Probability exceeds 1 | Overlapping cases double-counted | Check whether successful routes are disjoint |
Independent practice
- A fair die is rolled. Find P(even number).
- A fair coin is tossed four times. Find the probability of at least one head.
- A bag has 4 green and 3 yellow counters. Two are drawn without replacement. Find P(two green).
- Using the same bag, find P(exactly one green).
- Two fair dice are rolled. Find P(sum = 8).
- A machine produces a satisfactory item with probability 0.92 independently for each item. Find P(two satisfactory items).
- An event has probability 0.18. Find its expected frequency in 600 trials.
- A two-branch stage has probabilities 0.64 and x. If these are the only branches, find x.
Explained answers
1. The even outcomes are 2, 4 and 6, so P(even) = 3/6 = 1/2.
2. P(no heads) = (1/2)⁴ = 1/16. Therefore P(at least one head) = 15/16.
3. P(GG) = 4/7 × 3/6 = 2/7.
4. P(GY)=4/7×3/6=2/7. P(YG)=3/7×4/6=2/7. Total = 4/7.
5. Sum 8 occurs in (2,6),(3,5),(4,4),(5,3),(6,2), so probability = 5/36.
6. Independence gives 0.92² = 0.8464.
7. Expected frequency = 0.18×600 = 108.
8. Branches from the same node sum to 1, so x = 1−0.64 = 0.36.
Teaching sequence: representation before formula
Begin with language. Give the learner short phrases such as “both”, “either”, “at least one”, “exactly one” and “none”, and ask which outcomes they include. Then introduce possibility diagrams and simple trees before asking for combined numerical answers.
Next contrast replacement and non-replacement using the same bag. The changed branch probabilities should be explained before multiplication. Finally mix tree diagrams with complements so that the learner must choose between direct enumeration and a shorter opposite-event route.
Connect this guide to Set Language, Venn Diagrams and Counting for unions and intersections, and to Cumulative Frequency, Box Plots and Standard Deviation for the wider data-and-uncertainty system.
Final thought
The strongest probability solution makes the event structure visible. Once the sample space, branches and successful routes are clear, the arithmetic has somewhere reliable to live.
Represent the uncertainty first. Calculate only after the routes are clear.
Return to the Secondary Mathematics Hub.
