Loav · VCE Mathematical Methods · Units 3 and 4
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Probability becomes much easier when you decide which outcomes are still possible before calculating. This guide develops event notation, conditional probability, independence, tables and probability trees. You should already be comfortable with fractions, percentages and simple set diagrams. By the end, you should be able to translate a sentence into an event, identify the correct denominator, and justify why probabilities are multiplied or added.
A probability model describes a specified random experiment. Selecting one student uniformly from a list gives each student the same chance; selecting the first student to arrive does not automatically do so. Before using a count divided by a total, explain why the outcomes being counted are equally likely. Context supplies the model; arithmetic only works within that model.
1. Describe the events precisely
The sample space contains all possible outcomes. An event A is a collection of those outcomes. A′ is the complement, meaning A does not occur. The intersection A ∩ B means both events occur. The union A ∪ B means at least one occurs, including outcomes in both. In probability, “or” is normally inclusive unless the question explicitly says “exactly one”.
P(A′) = 1 − P(A)
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
The subtraction in the union rule corrects double counting: outcomes in the overlap were counted once in P(A) and again in P(B). The probability of exactly one event is P(A) + P(B) − 2P(A ∩ B). The event that neither occurs is the complement of the union. Drawing four regions, A only, B only, both and neither, is often more reliable than trying to remember several formulas at once.
Check that every region has nonnegative probability and that all four add to one. In particular, the overlap cannot exceed either individual probability. Impossible data should be identified rather than forced into a tree. A negative “probability” usually reveals a modelling or subtraction error.
2. Conditioning changes the reference group
P(A | B) means the probability of A given that B has occurred. The vertical bar is read “given”, not “divided by”. Once B is known, outcomes outside B are no longer candidates. Of the probability mass remaining in B, the part also in A is P(A ∩ B).
P(A | B) = P(A ∩ B)/P(B), provided P(B) > 0.
P(A ∩ B) = P(B)P(A | B) = P(A)P(B | A).
The condition belongs in the denominator. If a question says “among club members”, the denominator is the number of club members. P(A | B) and P(B | A) use the same intersection but usually different denominators. Neither can be reversed simply because the words sound similar. If P(B) = 0, the elementary ratio is undefined; do not report zero by cancelling or guessing.
3. Independence is a claim to test
Events are independent when knowing one occurred does not change the probability of the other. A test that also handles zero-probability events is P(A ∩ B) = P(A)P(B). When the relevant denominators are positive, you can instead compare P(A | B) with P(A), or P(B | A) with P(B). Show the two quantities you are comparing.
Mutually exclusive events cannot happen together, so their intersection has probability zero. If both have positive probability, they are dependent: knowing one occurred makes the other impossible. Independence describes unchanged probability; mutual exclusivity describes no overlap. They are different properties, even though both are sometimes casually called “separate”.
Independence is not established by different labels, different people or a sequence of trials. A shared influence can connect outcomes. Sampling without replacement usually changes later probabilities. With replacement and adequate random mixing, repeated draws can reasonably be modelled as independent. State assumptions when the wording gives a model rather than guaranteed physical independence.
Worked examples
Worked example 1: one table, two conditional probabilities
In an original survey model, 240 students include 90 bus users and 120 club members. Sixty are both. One of the 240 students is selected uniformly. Let B mean bus user and C mean club member.
- Complete the four regions: both 60; bus only 90 − 60 = 30; club only 120 − 60 = 60; neither 240 − 150 = 90.
- P(B | C) = 60/120 = 1/2. The reference group is the 120 club members.
- P(C | B) = 60/90 = 2/3. This time only 90 bus users are candidates.
- P(B ∩ C) = 60/240 = 1/4, whereas P(B)P(C) = (90/240)(120/240) = 3/16. Since these differ, B and C are not independent.
As a second check, P(B | C) = 0.5 exceeds P(B) = 0.375. Knowing club membership changes the bus probability. The survey shows an association within this model; it does not establish that joining a club causes bus use.
Worked example 2: update branches without replacement
A bag contains five gold tokens and three silver tokens. Two tokens are selected randomly without replacement. Find the probability of two gold tokens, at least one silver token, and a gold first token given a silver second token.
- For two gold tokens, multiply along the path: P(G1 ∩ G2) = (5/8)(4/7) = 5/14. There are only four gold tokens remaining after the first gold draw.
- At least one silver is the complement of two gold: 1 − 5/14 = 9/14. This avoids adding three separate paths.
- P(S2) = (5/8)(3/7) + (3/8)(2/7) = 3/8. Add the disjoint gold–silver and silver–silver paths.
- P(G1 | S2) = [(5/8)(3/7)]/(3/8) = 5/7.
The second draw still has an unconditional silver probability of 3/8 by symmetry, but its conditional probability depends on the first draw. Equal marginal probabilities do not prove independence.
4. Reverse a condition by rebuilding the reference group
A tree normally follows the order in which information is supplied. Multiply conditional probabilities along a complete path to obtain the probability of its intersection. Add complete paths only when they are disjoint. To reverse the condition, first find all paths satisfying the new condition, then compare the desired path with that total.
Worked example 3: identify a source after a failed check
A workshop receives 65% of its parts from line A and 35% from line B. Under a stated model, 4% of A’s parts and 10% of B’s parts fail inspection. A randomly selected part fails. What is the probability it came from B?
- Define F as failure. The supplied rates are P(F | A) = 0.04 and P(F | B) = 0.10. The question asks for P(B | F).
- P(F) = 0.65 × 0.04 + 0.35 × 0.10 = 0.061. Failure can arise through either line.
- P(B ∩ F) = 0.35 × 0.10 = 0.035.
- P(B | F) = 0.035/0.061 = 35/61 ≈ 0.5738.
Although B supplies fewer parts, it supplies about 57.38% of failed parts under this model. Answering 0.10 would confuse a failure rate within B with the fraction of failed parts originating from B. Retain unrounded intermediate values because the final division can magnify early rounding.
5. A reliable solution routine
Start by defining events in words. Translate the requested probability before inserting numbers. Choose a table when two classifications are measured on one group; choose a tree when probabilities change in a sequence. Label second-stage branches with their conditions, not just event names. Outgoing branches from each node should sum to one.
For “at least one”, check whether the complement is easier. For “given”, identify the reference group. For “independent”, compare probabilities rather than assert it from the story. Finish with an exact fraction when available, followed by an appropriate decimal if requested. A probability is dimensionless and must lie between zero and one; a percentage is the same quantity multiplied by 100.
6. Separate model probabilities from sample evidence
A statement that two events are independent belongs to a probability model. A finite survey can produce proportions that differ slightly even when an independent model is a reasonable approximation. Conversely, similar rounded percentages do not prove exact independence. In a mathematical question with specified exact probabilities, test the equality directly. In a data-based context, describe what the supplied numbers show without claiming more certainty than they support.
Also separate chronology from conditioning. The information after the vertical bar may be learned later than the event being asked about. In the token example, learning the second colour changes what can be inferred about the first draw; it does not physically change the earlier draw. Conditional probability is a calculation using available information.
Practice and reasoned answers
1. P(A) = 0.55, P(B) = 0.40 and P(A ∩ B) = 0.25. Find the probability of neither event and of exactly one.
The union is 0.55 + 0.40 − 0.25 = 0.70, so neither has probability 0.30. Exactly one removes the overlap twice from the sum: 0.55 + 0.40 − 0.50 = 0.45. The remaining probability, 0.25, is both; the three disjoint cases sum to one.
2. P(A ∩ B) = 0.18 and P(B) = 0.30. Find P(A | B). Can you find P(B | A)?
P(A | B) = 0.18/0.30 = 0.60. You cannot determine the reverse condition without P(A) or equivalent information. The intersection alone does not specify the size of the A reference group.
3. Independent events have P(A) = 0.7 and P(B) = 0.2. Find P(A ∩ B′).
P(A ∩ B) = 0.7 × 0.2 = 0.14. Split A into its B and B′ portions: P(A ∩ B′) = 0.7 − 0.14 = 0.56. Equivalently, independence extends to complements, giving 0.7 × 0.8.
4. A box contains four green and two red tiles. Two are drawn without replacement. Find the probability of different colours.
The two disjoint paths are green–red and red–green. Their probabilities are (4/6)(2/5) and (2/6)(4/5), so the answer is 16/30 = 8/15. Using a denominator of six on the second draw would incorrectly model replacement.
5. Two mutually exclusive events each have probability 0.3. Are they independent?
No. Their intersection probability is zero, while the product of their individual probabilities is 0.09. Once either occurs, the other becomes impossible. The positive individual probabilities make the difference decisive.
6. A printer uses tray A 80% of the time. Jam rates are 2% from A and 6% from tray B. Given a jam, find the probability tray A was used.
P(jam) = 0.8 × 0.02 + 0.2 × 0.06 = 0.028. The A-and-jam path is 0.016. Thus P(A | jam) = 0.016/0.028 = 4/7 ≈ 0.5714. Include both routes to a jam in the denominator.
7. P(A) = 0.4, P(B) = k and P(A ∪ B) = 0.7. Find k if A and B are independent.
Independence makes the intersection 0.4k. The union equation is 0.4 + k − 0.4k = 0.7, so 0.6k = 0.3 and k = 0.5. Check the overlap is 0.2 and the union is 0.4 + 0.5 − 0.2 = 0.7.
Learning check
Explain why conditioning changes a denominator. Draw a two-stage tree whose second-stage probabilities differ, and give an example of mutually exclusive events that are dependent. If you can calculate but cannot describe the reference group, return to the table example. Next, use these event rules to build and interpret a discrete probability distribution.
Curriculum and study route
Loav Mathematical Methods Units 3 and 4. These module divisions are a teaching sequence. Use the current VCAA Mathematics study design, Mathematical Methods Units 3 and 4, Area of Study 4, and the VCAA Mathematical Methods FAQ for course requirements. Scope checked 15 October 2026. All worked scenarios, questions and explanations in this guide are original.
