Conditional Probability & Distributions
Updating a probability on new information, and the shape of the normal curve.
The explanation
Conditional probability asks: given that B happened, how likely is A?
P(A|B) = P(A and B) / P(B)
Knowing B happened shrinks the world to just the B outcomes, so you divide by P(B) to rescale.
If knowing B tells you nothing about A, the events are independent and P(A|B) = P(A).
The normal distribution is the bell curve that describes heights, measurement errors and test scores. It is symmetric about the mean, and the empirical rule says roughly:
- 68% of values fall within 1 standard deviation of the mean
- 95% within 2
- 99.7% within 3
So a score 2 standard deviations above average beats about 97.5% of the group.
Conditional probability is defined as P(A|B) = P(A ∩ B)/P(B) for P(B) > 0, which rescales the sample space to B. Rearranging gives the multiplication rule P(A ∩ B) = P(B)P(A|B), and independence is the condition P(A|B) = P(A), equivalently P(A ∩ B) = P(A)P(B).
Bayes' theorem, P(A|B) = P(B|A)P(A)/P(B), inverts the conditioning and is the basis of diagnostic reasoning. Its most-cited consequence is that a test with high sensitivity can still yield mostly false positives when the base rate P(A) is low — an outcome that is counterintuitive but follows directly from the arithmetic.
The normal distribution N(μ, σ²) is symmetric and unimodal, with the empirical rule giving approximately 68/95/99.7% within one, two and three standard deviations. Standardising via z = (x − μ)/σ converts any normal variable to the standard normal, so a single table serves all cases.
The Central Limit Theorem explains the distribution's ubiquity: sample means approach normality as sample size grows regardless of the underlying population's shape, which is what licenses normal-based inference throughout statistics.
Worked example
In a class, 40% study Spanish, 25% study art, and 10% study both. Given a student studies Spanish, what is the probability they study art?
- P(art | Spanish) = P(both)/P(Spanish).
- = 0.10/0.40.
- = 0.25.
- Since P(art) = 0.25 too, the events are independent.
Answer: 0.25 — and the two subjects turn out to be independent here.
Common mistakes
- Dividing by P(A) rather than by the given event's probability P(B).
- Treating mutually exclusive events as independent. They are the opposite.