Secondary Mathematics · Worked Repair Guide 08
Probability does not predict a single future event with certainty. It describes how strongly a model supports different outcomes. A probability of 0.7 does not promise that the next ten trials will contain exactly seven successes. It says something about the structure of the chance model, not a guarantee about a short sequence.
This guide develops one central habit: define the sample space and the conditioning information before performing arithmetic. Many probability errors begin because the denominator is chosen too early, because replacement changes unnoticed, or because independent and mutually exclusive events are treated as the same idea.
All games, bags, cards and trial sequences below are original teaching examples. They are mathematical models, not gambling advice or predictions about real-world outcomes.
1. Probability begins with a sample space
A sample space lists the possible outcomes under the model. For one fair six-sided die, the sample space is {1,2,3,4,5,6}. For one fair coin, it is {H,T}.
If all outcomes in a finite sample space are equally likely, probability of an event equals the number of favourable outcomes divided by the number of possible outcomes.
For a fair die, P(even) = 3/6 = 1/2 because the favourable outcomes are 2, 4 and 6.
Entry check: for a fair die, P(number greater than 4)=2/6=1/3; P(number at most 4)=4/6=2/3; P(7)=0.
2. Probabilities live between zero and one
For any event A in an ordinary probability model, 0 ≤ P(A) ≤ 1. Zero means impossible under the model; one means certain under the model.
A result such as 1.3 or −0.2 cannot be a valid probability. That range check is one of the fastest ways to catch an arithmetic or modelling error.
Percentages can also represent probability. A probability of 0.35 is 35%. The fraction 7/20 represents the same number.
Keep the representation most useful for the calculation, and convert only when the question requires it.
3. The complement often turns a hard count into an easy one
The complement of event A means “A does not occur”. Since exactly one of A or not-A occurs, their probabilities sum to one:
P(not A) = 1 − P(A).
If P(rain)=0.3 in a stated model, P(no rain)=0.7.
Complements become especially useful for phrases such as “at least one”. If independent repeated trials make “none” easy to calculate, then P(at least one)=1−P(none).
Worked example: toss a fair coin twice. P(at least one head)=1−P(no heads)=1−1/4=3/4.
4. Two-stage sample spaces should preserve order when order matters
Tossing a coin twice gives ordered outcomes HH, HT, TH and TT. The middle two outcomes are different sequences even though each contains one head and one tail.
Thus P(exactly one head)=2/4=1/2.
A tree diagram provides another representation. Each branch corresponds to a stage, and multiplying probabilities along a complete path gives the probability of that ordered outcome when the branch probabilities are conditional on the path taken so far.
Do not compress HT and TH into one path unless the combined probability is calculated explicitly.
5. Addition combines disjoint routes to an event
If mutually exclusive outcomes lead to the same event, their probabilities can be added because they cannot occur together on one trial.
For a fair die, the event “roll 1 or 6” has probability 1/6 + 1/6 = 1/3.
For events that can overlap, simply adding P(A)+P(B) double-counts the overlap. The general relationship is:
P(A or B)=P(A)+P(B)−P(A and B).
For a fair die, let A be “even” and B be “greater than 3”. A={2,4,6}, B={4,5,6}, intersection={4,6}. So P(A or B)=3/6+3/6−2/6=4/6=2/3.
6. Multiplication follows a complete path
For two independent fair coin tosses, P(H then H)=1/2×1/2=1/4.
The multiplication is not a magic rule attached to the word “and”. It represents moving through successive stages of one path, using the probability appropriate at each stage.
If the second-stage probability changes after the first result, multiply by that changed conditional probability instead.
This distinction becomes essential in sampling without replacement.
7. Replacement determines whether the second-stage denominator changes
A bag contains 3 red and 2 blue counters. Draw one counter, replace it, then draw again. P(red then red)=3/5×3/5=9/25.
Without replacement, after drawing a red counter the bag contains 2 red and 2 blue counters. Therefore:
P(red then red)=3/5×2/4=3/10.
The second denominator changed from five to four because one counter was removed. The numerator also changed because the removed counter was red.
Repair habit: after each draw, write the new composition of the bag before writing the next probability.
8. Independence means one event does not change the probability of the other
Events A and B are independent when knowing that one occurred does not change the probability of the other. For independent events, P(A and B)=P(A)P(B).
Two separate fair coin tosses are independent in the standard model. The result of the first toss does not change the probability of heads on the second.
Two draws without replacement from the same small bag are generally not independent because the first draw changes the composition before the second.
Independence is therefore a statement about probability structure, not merely about events occurring at different times.
9. Mutually exclusive is not the same as independent
Mutually exclusive events cannot occur together. On one die roll, “even” and “odd” are mutually exclusive.
But they are not independent. If you know the outcome is even, the probability that it is odd becomes zero rather than remaining 1/2.
For two events with positive probabilities, mutually exclusive events cannot be independent because P(A and B)=0 while P(A)P(B)>0.
This contrast repairs a common vocabulary error. “Independent” does not mean “different” or “separate”. It has a precise probability meaning.
10. Conditional probability changes the denominator to the known condition
Suppose a fair die is rolled and you are told the result is greater than 3. The remaining possible outcomes are {4,5,6}. Given that information, the probability of an even result is 2/3 because 4 and 6 are favourable among three remaining equally likely possibilities.
The original denominator six is no longer the relevant sample space after the condition is imposed.
In notation, P(A|B) means probability of A given B. When P(B)>0:
P(A|B)=P(A and B)/P(B).
Use this extension only where it matches the learner’s course. The underlying idea—restrict the sample space to what is now known—remains useful even without formal notation.
11. Expected frequency is a model expectation, not a promised count
If an event has probability p and an experiment is repeated n times under stable identical conditions, the expected frequency is np.
If p=0.4 and n=200, the expected frequency is 80. This does not guarantee exactly 80 occurrences.
The result is a long-run model benchmark. Actual frequencies can vary from trial to trial.
Worked example: a model gives probability 0.15 of a certain outcome. Across 500 trials, expected frequency is 75. Observing 72 would not by itself prove the model false; observing 75 would not prove it true. Statistical model assessment requires more than matching one expectation exactly.
12. Experimental probability is an observed relative frequency
If an outcome occurs 38 times in 100 trials, the experimental probability or relative frequency is 0.38.
This is an observation from the sample, not automatically the true theoretical probability of the underlying process.
Repeating more trials can make relative frequency more stable in many ordinary models, but no finite sample guarantees exact equality with the theoretical probability.
Keep theoretical and experimental probabilities labelled. One comes from a mathematical model; the other from observed trial outcomes.
13. “At least”, “at most” and “exactly” define different events
For three coin tosses, “exactly two heads” means HHT, HTH or THH: probability 3/8 for fair independent tosses.
“At least two heads” includes exactly two or exactly three, so probability = 3/8 + 1/8 = 1/2.
“At most two heads” includes zero, one or two heads, which is the complement of three heads: 1−1/8=7/8.
Translate these phrases into included counts before calculating. The vocabulary determines the event set.
14. Counting paths prevents missing or double-counting outcomes
Suppose two fair dice are rolled. There are 36 equally likely ordered pairs. A sum of 7 occurs through six pairs: (1,6),(2,5),(3,4),(4,3),(5,2),(6,1). Therefore P(sum 7)=6/36=1/6.
The sums 2 through 12 are not equally likely. There are eleven possible sums, but they have different numbers of underlying ordered pairs.
A table or systematic list is often safer than assuming that distinct labels are equally likely.
Worked comparison: P(sum 2)=1/36 while P(sum 6)=5/36. The sample-space count, not the number printed as the sum, determines the probability.
15. Probability models depend on their assumptions
A statement such as “each die face has probability 1/6” belongs to a fair-die model. A loaded die can have different probabilities.
Likewise, independence may be an assumption of the model rather than something guaranteed by the mere existence of two trials.
When a school question states “fair”, “random” or “with replacement”, those words supply mathematical conditions. Read them as carefully as numerical values.
For real-world probability claims, the model, data source and assumptions matter. A numerical probability does not explain itself.
16. Capstone: two draws without replacement
A bag contains 4 red, 3 blue and 1 green counter. Two counters are drawn without replacement. Find the probability of two red counters, exactly one red counter, and at least one red counter.
Two red: 4/8×3/7 = 12/56 = 3/14.
Exactly one red has two routes. Red then non-red: 4/8×4/7 = 16/56. Non-red then red: 4/8×4/7 = 16/56. Total = 32/56 = 4/7.
At least one red can be calculated by complement. P(no red)=4/8×3/7=12/56=3/14. Therefore P(at least one red)=11/14.
Check: P(two red)+P(exactly one red)+P(no red)=3/14+8/14+3/14=1.
Notice that the two “exactly one red” path probabilities happen to be equal here, but they should still be represented as two ordered routes. In other compositions they need not be equal.
17. Independent practice
- For a fair die, find P(number less than 3).
- For a fair die, find P(not rolling a 6).
- Toss a fair coin twice. Find P(two tails).
- Toss a fair coin twice. Find P(exactly one tail).
- A bag has 5 red and 3 blue counters. Draw with replacement twice. Find P(red then blue).
- Use the same bag without replacement. Find P(red then blue).
- Using the bag in Question 6, find P(two red counters).
- Explain whether two draws without replacement are independent.
- On one fair die roll, are “even” and “greater than 4” mutually exclusive?
- For a fair die, find P(even or greater than 4).
- Given that a fair die result is greater than 2, find P(result is even).
- An event has probability 0.24. Find its expected frequency in 500 trials.
- An outcome occurs 47 times in 200 trials. Find its experimental probability.
- Toss a fair coin three times. Find P(exactly three heads).
- Toss a fair coin three times. Find P(at least one head).
- Roll two fair dice. Find P(sum 12).
- Roll two fair dice. Find P(sum 5).
- A bag contains 2 red and 3 blue counters. Two are drawn without replacement. Find P(at least one red).
18. Worked answers
1. 1/3. Favourable outcomes are 1 and 2 out of six equally likely faces.
2. 5/6. Use the complement of rolling a 6.
3. 1/4. Multiply 1/2×1/2 along the TT path.
4. 1/2. The ordered paths HT and TH each have probability 1/4.
5. 15/64. Replacement keeps the second probabilities at 5/8 and 3/8.
6. 15/56. After red is removed, 3 blue remain among seven counters.
7. 5/14. Compute 5/8×4/7.
8. Generally no. The first draw changes the composition and therefore the second-stage probability.
9. No. The outcome 6 belongs to both events, so they overlap.
10. 2/3. The union is {2,4,5,6}, four outcomes out of six.
11. 1/2. Given greater than 2, the remaining equally likely outcomes are {3,4,5,6}; even outcomes are 4 and 6.
12. 120. Expected frequency = 0.24×500.
13. 0.235. Compute 47/200.
14. 1/8. Only HHH works.
15. 7/8. Complement of no heads, which has probability 1/8.
16. 1/36. Only (6,6) gives sum 12.
17. 1/9. Sum 5 occurs through (1,4),(2,3),(3,2),(4,1): four out of 36.
18. 7/10. Complement of drawing two blue: 1−(3/5×2/4)=1−3/10.
19. Diagnose probability errors by the sample space they assume
Common errors include treating non-equally-likely labels as equal, forgetting that order creates multiple routes, failing to update a bag after no replacement, confusing mutually exclusive with independent, or using the original denominator after conditioning information is supplied.
A useful correction note states the changed sample space: “After one red is removed, four counters remain”, or “Given greater than 2, only four die outcomes remain”. That statement often fixes the arithmetic naturally.
Practice should vary the condition. Compare replacement with no replacement, “exactly one” with “at least one”, and independent trials with dependent draws. The contrast forces the learner to rebuild the probability structure.
20. Continue through the BTT Mathematics library
Return to the BTT Mathematics Hub. Use Averages, Spread and Data Interpretation for observed data summaries, Ratio, Percentage and the Correct Base for denominator discipline, and Graphs, Tables and Relationships for representing relationships visually.
The BTT Mathematical Lab is the diagnostic route when a sample-space or denominator error keeps returning.
21. Sources and scope
All probability experiments, sample spaces and trial counts in this guide are original teaching examples. Fairness, replacement and independence are stated model assumptions where used. Expected frequency is presented as a model expectation, not a guaranteed observed count.
For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match formal notation and extension sections to the learner’s actual subject level and school programme.