Probability & Bayes' Theorem
Probability is just "how many good outcomes out of all outcomes", written in the shortest way possible. Once you can see the two rules (add, multiply) as OR and AND, everything else is a variation.
ADD for "or". MULTIPLY for "and".
Every formula in this unit is one of those two, possibly with a condition attached. If you remember nothing else, this will get you most of the marks.
1. Terminology and Definition
| Term | Meaning |
|---|---|
| Random experiment | An activity with an uncertain result (roll a die, draw a card) |
| Sample space (S) | The set of all possible outcomes |
| Event (A) | Any subset of the sample space — a collection of outcomes you care about |
| Favourable outcomes | The elements of the event A |
| P(A) | Probability of A = favourable ÷ total |
| Complement Ā | Everything except A |
| Certain event | P = 1 (the whole sample space) |
| Impossible event | P = 0 (the empty set ∅) |
| Equally likely outcomes | Needed for the basic formula to be valid |
Question: two dice are thrown. What is the probability of getting a total of 7?
Bad way: list the 6 winning pairs. Good way:
Longer question: "probability of getting at least one six in 3 throws".
"at least one", "at least a", "not", "no", "none", "more than", "fewer than", "everything except", "impossible" → rewrite as 1 − P(the simpler opposite).
"Not getting a 6" is easier than "getting a 6"? Both are easy. But "at least one 6 in 10 throws" is horrible directly and trivial by complement. That is the whole point.
2. Mutually Exclusive vs Non-Mutually Exclusive
Two events are mutually exclusive if they cannot both happen at once — they share no outcome at all. Overlap = 0.
Adding P(A) + P(B) when A and B can both happen double counts the overlap. Always ask: "can this happen twice at once?" If yes, you need the general formula.
3. Dependent vs Independent Events
| Definition | Test | Example | |
|---|---|---|---|
| Independent | One event does not change the other's probability | P(A∩B) = P(A)·P(B) | Two dice. Rolling the first tells you nothing about the second. |
| Dependent (conditional) |
One event changes the other's probability | P(A∩B) ≠ P(A)·P(B) | Two cards from one deck. The first card changes what is left. |
- Can they happen together? Yes → not mutually exclusive, so subtract the overlap on the way in.
- Does the first change the second? No (replacement / separate trials) → independent, multiply the plain probabilities.
- Is it conditional? "Given that", "if we know", "suppose" → write P(B | A), never P(B).
4. Laws of Probability
Addition law
Question: two cards drawn from a 52-card deck. P(both are aces)?
Method 1 — sequential (cleanest):
Method 2 — with the addition law (shows the structure):
P(2nd is ace) = 4/52 (the second card is a different card, so do not use 3/51 here)
P(both) = 4/52 + 4/52 − P(1st ace AND 2nd ace)
= 8/52 − (4/52)(4/52) = 0.1538 − 0.00592 = 1/221 ✓
Multiplication law
Two results that always help
(b) Independent events, any number of them P(A₁ ∩ A₂ ∩ … ∩ Aₙ) = P(A₁)P(A₂)…P(Aₙ)
Question: a coin is tossed 4 times. What is the probability of getting exactly 3 heads?
Each toss is independent. The number 3 can land in any of the 4 positions (that is ⁴C 3 = 4 — this is where Unit 2 comes back):
5. Conditional Probability
Visually, you shrink the sample space down to B and then ask what fraction of that new space is A.
Question: two dice are thrown. Given that the sum is 8, what is P(both are even)?
Step 1 — shrink the space. Outcomes summing to 8: (2,6) (3,5) (4,4) (5,3) (6,2) → 5 outcomes.
Step 2 — count the good ones. Both even: (2,6), (4,4), (6,2) → 3 outcomes.
For any conditional question, write out the new sample space in full (a small table or list), cross out everything the condition rules out, and then count. It is slower but it is almost never wrong, and it shows the examiner you understand the definition.
6. Bayes' Theorem
Bayes flips a conditional probability backwards. It answers: "I already know the test result — now what is the real chance?"
Given: 1% of people have a disease (P(D) = 0.01). The test is 99% accurate (P(+|D) = 0.99 and P(−|D̄) = 0.99).
Step 1 — what we want.
Step 2 — find the pieces.
P(−|D̄) = 0.99 → P(+|D̄) = 0.01 (1% false alarms)
P(D) = 0.01, P(D̄) = 0.99
Step 3 — build the total for "positive" using the addition law.
= (0.99)(0.01) + (0.01)(0.99)
= 0.0099 + 0.0099 = 0.0198
Step 4 — divide.
- Write the target P(A | B) at the top. Find the two numbers on the right that you already know.
- Work out the denominator P(B) by splitting B into "A part" and "not-A part" and adding: P(B) = P(B|A)P(A) + P(B|Ā)P(Ā)
- Divide. Never leave a bare P(B) in the denominator — the examiner is looking for that split.
Mnemonic for the pattern: "target on top, split below, divide last."
- Independent events can still overlap — two dice can both show 6. Independent and mutually exclusive are different questions, and both can be true at once.
- P(A|B) ≠ P(B|A) except when P(A) = P(B). This is the classic trap in exams.
- "Mutually exclusive" is a stronger claim than "independent." Two events with P = 0 are independent and mutually exclusive. Positive-probability events cannot be both.
Self-check
Open to see the answers
1. One card is drawn from a standard deck. P(a heart OR a face card)?
P = 13/52 + 12/52 − 3/52 = 22/52 = 11/26
Note the −3/52: without it you would count J♥, Q♥, K♥ twice.
2. A bag has 3 red and 5 blue. Two drawn without replacement. P(both red)?
3. A factory: 60% of items come from machine A which makes 2% defective. Machine B makes 40%, 5% defective. A random item is defective — from which machine?
P(B ∩ def) = 0.40 × 0.05 = 0.020
P(def) = 0.012 + 0.020 = 0.032
P(A | def) = 0.012 / 0.032 = 0.375 = 37.5%
4. P(A) = 0.6, P(B) = 0.5, P(A∩B) = 0.2. What is P(A or B)? Are A and B independent?
Independent? P(A)P(B) = 0.6 × 0.5 = 0.30, but P(A∩B) = 0.20 → not independent