Probability becomes difficult when different outcomes are mixed together. A fraction may look reasonable even though its numerator counts sequences and its denominator counts groups, or the second draw uses the original bag contents after an item has been removed.
A list, table or tree can make the model visible. Choose the representation that shows the process clearly, then calculate from the outcomes it represents.
This guide uses original examples for O Level revision. Match practice questions to your syllabus and examination year. For wider planning, see Abeel’s O Level Mathematics guide.
Describe the random process first
Identify what is selected or observed, how many stages occur and whether anything changes between stages. A single die roll and two consecutive rolls are different experiments.
A bag draw with replacement differs from a draw without replacement. A fair coin differs from a coin whose probabilities are supplied by an experiment or model.
Write the assumptions given in the question. Do not silently assume fairness or independence simply because the objects are familiar.
A clear description determines the possible outcomes and their probabilities. Without it, counting can produce a neat answer for the wrong process.
Use equally likely outcomes when dividing counts
For a fair six-sided die, the six individual face results are equally likely. The probability of an even result is three qualifying faces out of six, or one half.
The descriptions “even” and “odd” happen to have equal probabilities here because each contains three faces. But outcome descriptions are not automatically equally likely merely because there are two of them.
For example, “a six” and “not a six” are two descriptions with probabilities one sixth and five sixths. Assigning one half to each would ignore the underlying face outcomes.
Before calculating favourable outcomes divided by total outcomes, check that the outcomes being counted are equally likely under the stated model.
List outcomes systematically
For two independent fair coin tosses, ordered outcomes are HH, HT, TH and TT. Each has probability one quarter under that model.
Exactly one head occurs in HT and TH, so its probability is two quarters, or one half. At least one head occurs in HH, HT and TH, so its probability is three quarters.
HT and TH are different ordered results because the first and second tosses are distinguished. Combining them into one outcome without adjusting its probability would distort the count.
Use a fixed listing order during revision. A systematic list helps prevent omissions and duplicated outcomes, especially when the question adds a condition.
A sample-space table helps with two stages
For two independent fair dice, put the first die’s results along one edge and the second die’s results along the other. The table contains thirty-six equally likely ordered pairs.
A total of seven occurs for (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1). The probability is six out of thirty-six, or one sixth.
The possible totals from two through twelve are not equally likely. A total of two has only one ordered pair, while seven has six.
Do not divide by eleven merely because there are eleven different total values. The table explains how many underlying equally likely outcomes contribute to each total.
Read a two-way table through its groups
Suppose sixty students are surveyed. Twenty-four enjoy reading, thirty enjoy sport, and twelve enjoy both. These are invented figures for practice.
The reading-only group contains twelve, the sport-only group eighteen, and eighteen students enjoy neither. The four groups total sixty.
The probability that a randomly selected student enjoys both is 12/60 = 1/5. The probability of reading or sport is (12 + 12 + 18)/60 = 42/60 = 7/10.
Check each group against the stated totals. A table is useful only if the overlap has been handled correctly rather than counted as an extra group outside the totals.
Correct for overlap when adding events
In the survey example, adding 24/60 and 30/60 counts the twelve students who enjoy both twice. Subtracting 12/60 once gives the correct union probability of 42/60.
When two events cannot happen together in the process being considered, they are mutually exclusive and there is no overlap to subtract.
For a single fair die roll, a two or a five is an example of mutually exclusive alternatives. Across multiple rolls, the event definitions would need to specify positions or sequences before exclusivity could be decided.
Ask whether one complete outcome can satisfy both descriptions. If it can, simple addition needs an overlap check.
Multiply along a complete tree path
A tree shows stages of a process. For independent fair coin tosses, each first-stage branch has probability one half and each second-stage branch also has probability one half.
The probability of heads then tails is 1/2 × 1/2 = 1/4. A complete path describes both events occurring in that order.
If several different complete paths satisfy the requested event, add their probabilities when those paths are mutually exclusive. Exactly one head has paths HT and TH, giving 1/4 + 1/4 = 1/2.
Do not add all branch labels across different levels and expect the total to be one. The alternatives leaving the same point total one when they cover all possibilities at that stage.
Update the tree after a draw without replacement
Suppose a bag contains five red and three blue counters. Two are drawn randomly without replacement. If the first is red, four red and three blue remain among seven counters.
The probability of two red counters is therefore 5/8 × 4/7 = 5/14. The second red probability is not still 5/8 because the bag has changed.
If the first is blue, five red and two blue remain. The second-stage branches must reflect that different first result.
Write the remaining counts beside each branch. This keeps the conditional information visible and prevents copying one second-stage fraction into every part of the tree.
Combine different orders for exactly one colour
For exactly one red in the two-draw bag example, the possible colour orders are red then blue and blue then red.
Their probabilities are 5/8 × 3/7 and 3/8 × 5/7. Adding gives 30/56 = 15/28.
These paths are mutually exclusive descriptions of the ordered draws, so their probabilities can be combined directly. Counting only one order would miss half of the qualifying sequences in this particular example.
Check all broad outcomes: two reds has probability 10/28, exactly one red 15/28, and two blues 3/28. Their sum is one, which confirms that this partition covers the process consistently.
Compare replacement with no replacement
If the first counter is returned and the bag is mixed before the second random draw, the original five-red, three-blue composition is restored.
Under the standard independent random-draw model with replacement, the probability of two reds is 5/8 × 5/8 = 25/64. Without replacement it was 5/14.
Both calculations multiply along a path. The distinction is whether the second probability changes after the first selection.
Read the replacement condition rather than inferring it from the fact that there are two draws. If the question supplies another dependence structure, use that structure instead of forcing either familiar bag model onto it.
Use complements when the opposite event is easier
The probability of at least one head in two independent fair coin tosses is one minus the probability of no heads. No heads means TT, with probability one quarter, so the answer is three quarters.
A complement must be the exact opposite event. The complement of “at least two successes” includes zero and one success, not merely zero.
Describe the opposite in words before calculating. Then check that the original event and the complement cannot occur together and jointly cover every possible result.
This method can save work, but it is not a reason to ignore the random process or its assumptions. The complementary event’s probability still needs a valid calculation.
Restrict the group when information is given
In the sixty-student survey, suppose you know that the selected student enjoys sport. The relevant group is now the thirty sport students rather than all sixty students.
Twelve of those thirty also enjoy reading, so the probability of reading given sport is 12/30 = 2/5.
The overlap remains twelve, but the denominator changes with the condition. Reversing the condition would compare the overlap with the twenty-four reading students instead.
A two-way table makes this restriction easy to see. Identify the group named by the information already given before choosing the denominator.
Distinguish experimental frequency from a model guarantee
If an outcome appears eighteen times in fifty trials, its relative frequency is 18/50 = 0.36. That observed proportion can inform an estimate, but it does not guarantee the same fraction in the next fifty trials.
An expected frequency is also not a promise about one batch. If a model assigns probability 0.3 to an event, forty trials have expected frequency twelve, but an actual result can differ.
Read whether the question gives theoretical probabilities, observed data or a requested estimate. Use the appropriate interpretation rather than treating every fraction as an exact permanent property.
Record the event, process and chosen representation in your working. Use Abeel’s Cambridge past-paper guide and O Level revision plan for independent practice and later correction.
Explore Abeel Eschool and the course catalogue for current learning options. Ask through admissions and FAQs, bringing your outcome list or tree so the difficult step is clear.
Frequently asked questions
Are all possible descriptions equally likely?
No. Check the underlying outcomes and random process. “A six” and “not a six” have different probabilities for a fair die.
Why do I multiply along a tree path?
A complete path requires its successive events to occur together. Use the appropriate probability at each stage.
Why do second-draw probabilities change without replacement?
Removing an item changes the remaining total and may change the relevant colour count. Each branch must reflect what happened first.
Do I add both orders for exactly one red?
Include every qualifying order, such as red-blue and blue-red, when they are distinct possible paths in the stated process.
Does an expected frequency have to happen exactly?
No. It describes average behaviour under the model, not a guaranteed result for one set of trials.
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