Equally likely outcomes & counting
When every outcome is equally likely, P(A) = |A|/|Ω| and probability reduces entirely to counting. The only real decision is whether order matters: if you count the numerator with combinations, you must count the denominator the same way. Mixing ordered and unordered counts is what makes these questions go wrong.
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Method: how to approach it
The order below is what examiners expect to see, and each step carries its own marks.
- Decide whether order mattersCommittees and hands are unordered; sequences and rankings are ordered. Commit to one and use it throughout.
- Count the total outcomesUsually C(n,k) for a selection or nᵏ for a sequence with repetition.
- Count the favourable outcomesBreak the event into independent choices and multiply, or use complements when "at least" appears.
- Divide and simplifyCancel factorials before multiplying out — the numbers stay small and exact.
Worked example
Five cards are dealt from a standard pack. Find the probability of exactly two aces.
- Total hands: C(52,5) = 2,598,960.
- Choose 2 aces from 4: C(4,2) = 6.
- Choose the other 3 cards from the 48 non-aces: C(48,3) = 17,296.
- Favourable: 6 × 17,296 = 103,776.
Answer. 103,776 / 2,598,960 ≈ 0.0399, about a 4% chance.
Where marks get dropped
These are the specific errors that cost credit on equally likely outcomes & counting questions — QED's rubric penalises each of them separately.
- Counting the numerator as ordered and the denominator as unordered. The two must be consistent.
- Forgetting to restrict the remaining choices. The other three cards must come from the 48 non-aces, or you count hands with three aces too.
- Assuming outcomes are equally likely when they are not — totals on two dice being the standard trap.
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Equally likely outcomes & counting — frequently asked questions
When can I use |A|/|Ω|?
Only when every outcome in Ω is equally likely. Choose the sample space so this holds, even if it means a larger Ω.
Does order matter for a card hand?
No — a hand is a set. But you may count with order in both numerator and denominator and the factors cancel, giving the same answer.
Why is the birthday problem surprising?
Because the complement counts PAIRS, and 23 people give 253 pairs. Computing P(no shared birthday) as a product of shrinking fractions drops below 1/2 at 23.
The rest of Probability
Events, conditional probability, random variables. Each subtopic below has its own method, worked example and mark-losing traps.
- 1Sample spaces & events
- 2Conditional probability & independence
- 3Bayes’ theorem
- 4Random variables & expected value
- 5Variance & standard deviation
- 6Binomial & uniform distributions
- 7Law of total probability & tree diagrams
- 8Geometric & Poisson distributions
- 9Joint, marginal & conditional distributions
- 10Markov & Chebyshev bounds
- 11Equally likely outcomes & counting
- 12Linearity of expectation
- 13Indicator random variables
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