Mathematics Tutor Singapore | Sec 4 Probability

Mathematics Tutor Singapore | Secondary 4 probability questions often go wrong before the calculation begins. A child may multiply two fractions correctly but forget that “one red and one blue, in either order” describes two different paths through the experiment. The first teaching task is defining the complete event.

Did you know that mutually exclusive events and independent events describe two very different relationships? Mutually exclusive events cannot occur together. Independent events do not change one another’s probability. Confusing those definitions is one of the most common ways to lose marks in tree diagrams and combined probability.

This original G3 E-Maths tuition guide explains outcomes, probability trees, fair dice and coins, with- and without-replacement draws, complements and the addition rule. It is grounded in the 2027 Singapore-Cambridge SEC G3 Mathematics syllabus and prioritises explaining the model before pressing buttons.

What probability means

Probability is a numerical measure of chance from zero to one. An impossible event has probability zero and a certain event has probability one. A probability of one-half means a theoretical chance of one in two under a specified model; it does not guarantee that every two actual trials contain one success.

For a fair six-sided die, each face has probability one-sixth. The even faces are 2, 4 and 6, so their combined probability is three-sixths or one-half. This favourable-over-total counting method works because all six basic outcomes are assumed equally likely.

Not all experiments have equally likely elementary outcomes. A real weighted die may favour some faces. The question must supply fair conditions or relevant numerical probabilities before the method can be justified.

Identify what is being randomly selected or generated.
Write the possible outcomes before identifying the event.
Mark whether order matters.
Check whether probabilities change between stages.
List every valid event path.
Verify that all complete path probabilities together total one.

Two independent fair coin tosses

A coin tossed twice has four equally likely ordered sequences: HH, HT, TH and TT, under the independent fair-coin model. The two sequences HT and TH are different experimental outcomes even though each contains one head and one tail.

Event	Qualifying outcomes	Probability
Exactly one head	HT, TH	2/4=1/2
At least one head	HH,HT,TH	3/4
Both heads	HH	1/4
No head	TT	1/4

A complete event may contain more than one ordered outcome.

A student who lists only HH, HT and TT has silently combined HT and TH into one outcome. Those three categories would not have equal probabilities. Treating them as three equally likely results makes later answers unreliable.

How a probability tree works

Draw the first branch to H and T, each with probability 1/2. From each one, draw a second branch to H and T, again with probability 1/2. The complete paths are HH, HT, TH and TT. Their probabilities are each 1/2×1/2=1/4.

A tree records the order of events and the branch probability given the earlier outcome. If the second experiment has been affected by the first, its branch labels must change accordingly. This is why a tree is valuable for sampling from a bag without replacement.

First stage	Second stage	Path probability
H, 1/2	H, 1/2	1/4
H, 1/2	T, 1/2	1/4
T, 1/2	H, 1/2	1/4
T, 1/2	T, 1/2	1/4

Path probabilities multiply and all complete paths sum to one.

Multiply down a path, add disjoint valid paths

For a specified sequence such as head then tail, multiply probabilities along the path. For an event containing either head-tail or tail-head, add those disjoint path probabilities. In the two-coin example, 1/4+1/4=1/2 for exactly one head.

I ask students to write “this path” or “these paths” beside their calculation. It prevents them from using the probability of one sequence as the answer to an event that includes several sequences.

Be careful with the slogan “and means multiply, or means add”. It is useful only with the correct probabilities and conditions. In particular, an ordinary addition of overlapping event probabilities double-counts their shared outcomes.

Mutually exclusive and independent events

Independent events do not alter each other’s chances. For instance, a fair coin toss and an independent die roll satisfy P(head and even)=P(head)P(even)=1/2×1/2=1/4.

Mutually exclusive events cannot both happen in the same trial. On one die roll, being even and being odd are mutually exclusive. Because both events have positive probability and their joint event is impossible, they cannot also be independent.

Events	Can occur together?	Classification
Coin head and independent die even	Yes	Independent, not mutually exclusive
Die even and die odd	No	Mutually exclusive; not independent
Die even and die greater than four	Yes, at face six	Neither mutually exclusive nor independent

Independence describes unchanged chance, while exclusivity describes no overlap.

The definitions should come before the algebra. Two positive-probability events with no joint outcome have intersection probability zero, whereas independence would require the product of their positive probabilities, which cannot equal zero.

Worked example: red and blue without replacement

A bag contains three red and two blue identical balls. Draw two in succession without replacement. Find the probability of one of each colour in either order. We need both red-blue and blue-red paths.

Red then blue has probability 3/5×2/4=3/10. Blue then red has probability 2/5×3/4=3/10. Since the two ordered paths are disjoint, the total probability is 3/5.

Path	First chance	Second chance	Joint chance
Red, blue	3/5	2/4	3/10
Blue, red	2/5	3/4	3/10
Either path	Add results	Two paths	3/5

The denominator for the second draw changes from five balls to four.

If the learner reports 3/10, they have identified one valid path but missed the alternative order. If they use 2/5 for the second colour after drawing red, they have failed to update the number of balls in the bag.

What changes when we replace the first ball?

With replacement, the bag composition is restored before the next random draw. The second draw then again has red probability 3/5 and blue probability 2/5 under the usual independently random sampling model.

The probability of two reds becomes (3/5)²=9/25, instead of (3/5)(2/4)=3/10 without replacement. Exactly one red and one blue with replacement has two paths, each 3/5×2/5=6/25, giving 12/25.

Event	Without replacement	With replacement
Two red	3/10	9/25
One of each in any order	3/5	12/25

One word—replacement—can change the entire probability tree.

The useful lesson is not that one sampling method always produces a larger answer. It is that the second-stage probabilities must reflect what actually happened during the first stage.

Probability of ‘or’ when events overlap

Roll one fair six-sided die. Let A be even faces {2,4,6}; let B be faces greater than four {5,6}. Their union contains {2,4,5,6}. So P(A or B)=4/6=2/3.

Adding 3/6+2/6 would give 5/6, which double-counts face six. The correct addition rule is P(A or B)=P(A)+P(B)−P(A and B). Here it yields 3/6+2/6−1/6=4/6.

A child may understand the fraction arithmetic but not why the subtraction is necessary. Listing the sets explicitly is often a clearer explanation than presenting the formula alone.

Use a complement for ‘at least one’

Toss a fair coin independently three times. The event “at least one head” includes one, two or three heads. The complement is no heads, which is the single sequence TTT. Its probability is (1/2)³=1/8, so the required probability is 1−1/8=7/8.

The complement strategy works because the desired event and its complement are disjoint and together cover the complete sample space. A useful verification is to list all eight ordered three-toss sequences and observe that exactly one has no head.

An independent dice example

Two distinguishable fair six-sided dice have thirty-six equally likely ordered outcomes. A sum of seven occurs in (1,6),(2,5),(3,4),(4,3),(5,2),(6,1), giving six ways and probability 1/6.

A sum of at least ten includes sums ten, eleven and twelve. These occur in three, two and one ordered ways respectively. Hence the event also has six favourable ways out of thirty-six, or one-sixth.

Do not assume the possible sums from two through twelve are equally likely. Seven has six ordered representations whereas two has only (1,1). A list of eleven different sums is not an equally likely sample space.

What the model assumes

The classroom calculations assume fair coins, fair dice, known bag contents and random draws. Real devices may not be perfectly fair, and experimental frequencies can fluctuate around their theoretical probabilities.

For example, seven heads in ten tosses do not prove a coin’s true head probability is 0.7; the small observed sample may deviate from the underlying probability. The exam question normally supplies the intended model, and the student should work within its stated assumptions.

Eight frequent Secondary probability errors

First error	Likely misconception	Next diagnostic question
Lists only HH,HT,TT for two tosses	Order ignored	Where is TH?
Computes RB but omits BR	Alternative event path omitted	What does ‘either order’ include?
Second red probability stays 3/5 without replacement	Bag contents not updated	How many red and total remain?
Calls even and odd independent	Definitions confused	Can both occur?
Adds overlapping events without subtracting intersection	Double-counting	Which face belongs to both sets?
Uses exactly one for ‘at least one’	Event wording misunderstood	Does two successes also qualify?
Treats sums 2–12 on two dice as equally likely	Underlying outcomes ignored	Count ordered die pairs
Produces a probability greater than one	Inconsistent outcome accounting	Can a chance exceed certainty?

Fix the event definition and probability model before the numerical calculation.

In tuition I look at the pupil’s first original tree or sample-space list. The wrong number alone does not show whether the issue is missing a path, failing to update a denominator or multiplying accurately but adding the wrong alternatives.