Conditional Probability

Conditional probability changes the reference population. For events $A$ and $B$ in a probability space, with $P(B)>0$,

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)}.$$

This definition is the algebraic base for Bayes’ theorem, diagnostic tests, classifier calibration, and likelihood calculations in hypothesis testing.

Intuition

Conditioning discards outcomes outside $B$ and renormalizes the remaining mass to one. The event $A$ is then evaluated only inside that narrowed sample space. This is why $P(A\mid B)$ and $P(B\mid A)$ can be very different.

Worked computation

The simulation below repeats the disease-testing base-rate scenario from Bayes’ theorem and compares the simulated conditional probability with the exact calculation.

import numpy as np
 
rng = np.random.default_rng(20260711)
N = 500000
disease = rng.random(N) < 0.01
positive = np.where(disease, rng.random(N) < 0.99, rng.random(N) < 0.05)
exact = (0.99 * 0.01) / (0.99 * 0.01 + 0.05 * 0.99)
print("sim_P(disease|positive)", round(disease[positive].mean(), 4))
print("exact", round(exact, 4), "positive_rate", round(positive.mean(), 4))

Observed output:

sim_P(disease|positive) 0.1646
exact 0.1667 positive_rate 0.0596

The simulation gives $P(disease\mid positive)\approx 0.1646$, close to the exact value 0.1667. Even with 99 percent sensitivity, most positives are false positives because the positive rate is only about 0.0596 and the false-positive term is applied to the much larger healthy group.

Caveats

Conditioning on a post-treatment event, selected sample, or model flag can introduce selection bias. Independence is a special claim: $P(A\mid B)=P(A)$, not the default.

References

• ProbabilityCourse: Conditional Probability
• OpenStax Introductory Statistics 2e, Chapter 3 introduction

Nav

Section — Probability and Statistics

← Expectation and Variance Bayes Theorem →