Probability Spaces

Quantum mechanics doesn't predict what you'll measure — it predicts the probabilities. Before we can read the Born rule honestly, we need the language those probabilities are written in: the probability space $(\Omega, \mathcal F, \mathbb P)$.

Learning Objectives

By the end of this lesson you will be able to:

1. State the Kolmogorov axioms and define a probability space $(\Omega, \mathcal F, \mathbb P)$, including why we need a $\sigma$-algebra of events.
2. Compute conditional probabilities and test independence of events from first principles.
3. Apply the law of total probability and derive Bayes' theorem, then use it to perform a Bayesian update.
4. Connect these objects to quantum measurement statistics (outcomes, the Born rule — forward-referenced to Term 1).
5. Estimate an event probability by Monte-Carlo simulation and verify Bayes' theorem numerically in NumPy.

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Intuition

A random experiment has some set of possible outcomes. Flip a coin: $\{H, T\}$. Roll a die: $\{1,\dots,6\}$. Measure a qubit prepared in $|+\rangle$ in the computational basis: $\{0, 1\}$, each with probability $\tfrac12$. The whole apparatus of probability theory is the bookkeeping that lets us assign consistent numbers to combinations of outcomes ("the die is even", "the qubit reads 0 twice in a row") and to update those numbers when we learn something new.

Three ingredients suffice:

• $\Omega$ — the sample space. Every possible elementary outcome, exactly once.
• $\mathcal F$ — the event space. The collections of outcomes we're allowed to ask about. An event is a subset of $\Omega$; "the die is even" is the event $\{2,4,6\}$.
• $\mathbb P$ — the probability measure. A rule assigning each event a number in $[0,1]$, obeying a short list of consistency axioms.

For finite $\Omega$ you could take $\mathcal F$ to be all subsets and never think about it again. The $\sigma$-algebra machinery only earns its keep on infinite/continuous sample spaces (e.g. a uniform real number in $[0,1]$), where you cannot consistently assign a probability to every subset. We set the machinery up carefully now because the same structure — outcomes, a sample space, a measure — is exactly what a quantum measurement produces. The difference is only in where the numbers come from: classically they're given; quantum-mechanically the Born rule computes them from a state vector. Same scaffolding, different engine.

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Theory

Sample space and events

The sample space $\Omega$ is the set of all elementary outcomes of an experiment. An event is a subset $A \subseteq \Omega$; we say "$A$ occurs" if the realized outcome $\omega \in A$. Set operations carry their logical meaning:$A \cup B$is "$A$or$B$",$A \cap B$is "$A$ and $B$", $A^c = \Omega \setminus A$ is "not $A$".

The event $\sigma$-algebra

We do not always want (or get to choose) every subset as an event. The admissible events form a $\sigma$-algebra (also "$\sigma$-field"): a collection $\mathcal F \subseteq 2^{\Omega}$ such that

$$\text{(i)}\ \ \Omega \in \mathcal F, \qquad \text{(ii)}\ \ A \in \mathcal F \Rightarrow A^c \in \mathcal F, \qquad \text{(iii)}\ \ A_1, A_2, \dots \in \mathcal F \Rightarrow \bigcup_{i=1}^{\infty} A_i \in \mathcal F .$$

That is: it contains the whole space, and is closed under complement and countable union (hence, by De Morgan, under countable intersection). The closure under countable operations — not just finite — is exactly what lets us talk about limits like "the long-run frequency converges", which we will need for the Law of Large Numbers in the next lesson.

Why not just use all subsets? For finite or countable $\Omega$ you can, and we usually take $\mathcal F = 2^\Omega$. But for $\Omega = [0,1]$ with the uniform "length" measure, Vitali's theorem shows there is no way to assign a consistent length to every subset. The standard choice is the Borel $\sigma$-algebra generated by the intervals. The $\sigma$-algebra is the precise statement of "which questions are answerable."

The Kolmogorov axioms

A probability measure is a function $\mathbb P : \mathcal F \to [0,1]$ satisfying Kolmogorov's three axioms [CT, §2]:

$$\textbf{(K1) Non-negativity:}\quad \mathbb P(A) \ge 0 \ \ \text{for all } A \in \mathcal F .$$$$\textbf{(K2) Normalization:}\quad \mathbb P(\Omega) = 1 .$$$$\textbf{(K3) Countable additivity:}\quad \mathbb P\!\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} \mathbb P(A_i) \quad\text{for pairwise disjoint } A_i\ (A_i \cap A_j = \varnothing,\ i\neq j).$$

The triple $(\Omega, \mathcal F, \mathbb P)$ is a probability space. Everything else is a theorem.

Basic consequences (each a short derivation from K1–K3):

Probability of the empty set. Take $A_1 = \Omega$, $A_2 = A_3 = \dots = \varnothing$. These are disjoint and union to $\Omega$, so by (K3) and (K2),

$$1 = \mathbb P(\Omega) = \mathbb P(\Omega) + \sum_{i\ge 2}\mathbb P(\varnothing) \;\Rightarrow\; \sum_{i\ge 2}\mathbb P(\varnothing) = 0 \;\Rightarrow\; \mathbb P(\varnothing) = 0 ,$$

using non-negativity (K1) to force each term to zero.

Complement rule. $A$ and $A^c$ are disjoint with $A \cup A^c = \Omega$, so

$$\mathbb P(A) + \mathbb P(A^c) = \mathbb P(\Omega) = 1 \;\Rightarrow\; \mathbb P(A^c) = 1 - \mathbb P(A).$$

Monotonicity. If $A \subseteq B$, write $B = A \,\sqcup\, (B \setminus A)$ (disjoint union). Then $\mathbb P(B) = \mathbb P(A) + \mathbb P(B\setminus A) \ge \mathbb P(A)$ by (K1). In particular $\mathbb P(A) \le \mathbb P(\Omega) = 1$.

Inclusion–exclusion (two sets). Decompose into disjoint pieces $A\cup B = (A\setminus B)\sqcup(A\cap B)\sqcup(B\setminus A)$:

$$\mathbb P(A \cup B) = \mathbb P(A) + \mathbb P(B) - \mathbb P(A \cap B).$$

Conditional probability

Learning that event $B$ occurred shrinks the world to $B$ and renormalizes. For $\mathbb P(B) > 0$, the conditional probability of $A$ given $B$ is defined as

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

This is a definition, but a principled one: fix $B$ with $\mathbb P(B)>0$ and check that $A \mapsto \mathbb P(A\mid B)$ is itself a probability measure on $(\Omega, \mathcal F)$. Indeed $\mathbb P(A\mid B)\ge 0$; $\mathbb P(\Omega \mid B) = \mathbb P(B)/\mathbb P(B) = 1$; and for disjoint $A_i$, since the $A_i \cap B$ are also disjoint, countable additivity of $\mathbb P$ pushes through the division. So conditioning yields a new probability space supported on $B$ — exactly the "shrink and renormalize" picture.

Rearranging gives the multiplication rule:

$$\mathbb P(A \cap B) = \mathbb P(A\mid B)\,\mathbb P(B) = \mathbb P(B \mid A)\,\mathbb P(A).$$

Quantum foreshadowing. Sequential measurements are conditional probability in disguise. If a qubit measured in the computational basis reads $0$, a second computational-basis measurement reads $0$ with certainty: $\mathbb P(\text{2nd}=0 \mid \text{1st}=0) = 1$. The measurement collapse of Term 1 is precisely the "shrink to $B$ and renormalize" operation, with the Born rule supplying the numbers.

Independence

Events $A$ and $B$ are independent if knowing one tells you nothing about the other:

$$A \perp B \;\iff\; \mathbb P(A \cap B) = \mathbb P(A)\,\mathbb P(B).$$

Equivalently, when $\mathbb P(B)>0$, $\mathbb P(A\mid B) = \mathbb P(A)$. Note independence is a statement about the measure, not about the outcomes being "unrelated" in any physical sense — and it is not the same as disjointness. Two disjoint events with positive probability are maximally dependent: if one occurs the other cannot, so $\mathbb P(A\cap B)=0 \ne \mathbb P(A)\mathbb P(B)$.

For a collection $\{A_i\}$, mutual independence requires $\mathbb P\!\big(\bigcap_{i\in S} A_i\big) = \prod_{i\in S}\mathbb P(A_i)$ for every finite subset $S$ — strictly stronger than pairwise independence.

Law of total probability

Let $\{B_1, B_2, \dots\}$ be a partition of $\Omega$: pairwise disjoint with $\bigcup_i B_i = \Omega$ and each $\mathbb P(B_i) > 0$. Any event $A$ splits along the partition, $A = \bigsqcup_i (A \cap B_i)$, so by countable additivity and the multiplication rule,

$$\mathbb P(A) \;=\; \sum_i \mathbb P(A \cap B_i) \;=\; \sum_i \mathbb P(A \mid B_i)\,\mathbb P(B_i) .$$

This is the workhorse for "average over the hidden cause." It says the marginal probability of $A$ is the weighted average of its conditional probabilities, weighted by how likely each cause $B_i$ is.

Bayes' theorem — derivation

Bayes' theorem inverts a conditional. Start from the two ways of writing the joint probability,

$$\mathbb P(B_j \cap A) = \mathbb P(A \mid B_j)\,\mathbb P(B_j) = \mathbb P(B_j \mid A)\,\mathbb P(A),$$

solve for the quantity we want, and substitute the law of total probability for the denominator:

$$\boxed{\;\mathbb P(B_j \mid A) = \frac{\mathbb P(A \mid B_j)\,\mathbb P(B_j)}{\mathbb P(A)} = \frac{\mathbb P(A \mid B_j)\,\mathbb P(B_j)}{\sum_i \mathbb P(A \mid B_i)\,\mathbb P(B_i)}\;}$$

The names: $\mathbb P(B_j)$ is the prior, $\mathbb P(A\mid B_j)$ the likelihood, $\mathbb P(B_j\mid A)$ the posterior, and the denominator $\mathbb P(A)$ the evidence (or marginal likelihood) — a normalizing constant ensuring the posteriors over $j$ sum to one. Bayes' theorem is the rule for updating beliefs in light of data, and it underpins the statistical inference we'll do on measurement counts throughout the program.

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Worked Examples

Example 1 — Medical test (a Bayesian update)

A disease has prevalence $\mathbb P(D) = 0.01$. A test is $99\%$ sensitive ($\mathbb P(+ \mid D) = 0.99$) and $95\%$ specific ($\mathbb P(- \mid D^c) = 0.95$, so the false-positive rate is $\mathbb P(+ \mid D^c) = 0.05$). You test positive. What is $\mathbb P(D \mid +)$?

The partition is $\{D, D^c\}$. Law of total probability for the evidence:

$$\mathbb P(+) = \mathbb P(+\mid D)\mathbb P(D) + \mathbb P(+\mid D^c)\mathbb P(D^c) = (0.99)(0.01) + (0.05)(0.99) = 0.0099 + 0.0495 = 0.0594.$$

Bayes' theorem:

$$\mathbb P(D \mid +) = \frac{\mathbb P(+\mid D)\,\mathbb P(D)}{\mathbb P(+)} = \frac{0.0099}{0.0594} \approx 0.1667 .$$

A positive result on a very accurate test still leaves only a $\sim 17\%$ chance of disease — because the prior is tiny, the base rate dominates. This base-rate effect is the single most common error in applied probability; Bayes' theorem makes it quantitative.

Example 2 — Measurement-outcome analogy (sequential qubit reads)

Model a noisy detector classically. A qubit is prepared so that an ideal computational-basis measurement would read $0$ with probability $q = \mathbb P(\text{ideal}=0) = 0.7$ and $1$ with probability $0.3$. The detector is imperfect: given the ideal outcome it reports correctly with probability $0.9$ and flips with probability $0.1$ (a readout error, exactly the kind we'll mitigate in Term 4). You read a reported value $R = 0$. What is the probability the ideal outcome was $0$?

Partition by the ideal outcome $\{I=0, I=1\}$. Likelihoods: $\mathbb P(R=0\mid I=0)=0.9$, $\mathbb P(R=0\mid I=1)=0.1$. Evidence:

$$\mathbb P(R=0) = (0.9)(0.7) + (0.1)(0.3) = 0.63 + 0.03 = 0.66.$$

Posterior:

$$\mathbb P(I=0\mid R=0) = \frac{(0.9)(0.7)}{0.66} = \frac{0.63}{0.66} \approx 0.9545 .$$

Reading $0$ raises our confidence that the ideal outcome was $0$ from the prior $0.7$ to $\approx 0.95$. This is precisely the inference performed by readout-error mitigation: invert the known noise (likelihood) to recover the underlying distribution from observed counts.

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Hands-on (Python)

We do two things: (1) estimate an event probability by Monte-Carlo simulation and watch the estimate converge, and (2) verify Bayes' theorem by simulating the joint experiment and comparing empirical conditional frequencies against the analytic posterior.

import numpy as np

rng = np.random.default_rng(0)  # seed for reproducible runs

# -------------------------------------------------------------------------
# (1) Monte-Carlo estimate of an event probability.
# Experiment: roll two fair six-sided dice. Event A = "sum >= 10".
# True value: outcomes {(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)} -> 6/36 = 1/6.
# -------------------------------------------------------------------------
def estimate_event_prob(n_trials, rng):
    d1 = rng.integers(1, 7, size=n_trials)   # uniform on {1,...,6}
    d2 = rng.integers(1, 7, size=n_trials)
    A = (d1 + d2) >= 10                       # boolean array: did A occur?
    return A.mean()                           # fraction of trials in A = P_hat(A)

true_p = 6 / 36
for n in (100, 1_000, 100_000, 1_000_000):
    p_hat = estimate_event_prob(n, rng)
    print(f"n={n:>9}: P_hat(A)={p_hat:.5f}  |error|={abs(p_hat-true_p):.5f}")

# The error shrinks like 1/sqrt(n) -- the standard-error scaling we DERIVE in
# the next lesson and that later governs how many measurement "shots" a QPU run
# needs for a target precision.

# -------------------------------------------------------------------------
# (2) Verify Bayes' theorem by simulation (the medical-test example).
# We simulate the JOINT experiment (disease status, test result), then read off
# the empirical posterior P(D | +) and compare to the analytic value.
# -------------------------------------------------------------------------
P_D       = 0.01    # prior prevalence
sens      = 0.99    # P(+ | D)
fpr       = 0.05    # P(+ | not D)

def analytic_posterior(P_D, sens, fpr):
    evidence = sens * P_D + fpr * (1 - P_D)         # law of total probability
    return (sens * P_D) / evidence                  # Bayes' theorem

def simulate_posterior(n, rng, P_D, sens, fpr):
    has_disease = rng.random(n) < P_D               # sample D ~ Bernoulli(P_D)
    # P(+ | status): sensitivity if diseased, false-positive rate otherwise
    p_positive = np.where(has_disease, sens, fpr)
    tested_pos = rng.random(n) < p_positive         # sample test result
    # Empirical P(D | +) = (# diseased AND positive) / (# positive)
    return (has_disease & tested_pos).sum() / tested_pos.sum()

post_exact = analytic_posterior(P_D, sens, fpr)
post_sim   = simulate_posterior(2_000_000, rng, P_D, sens, fpr)
print(f"\nP(D | +) analytic   = {post_exact:.5f}")
print(f"P(D | +) simulated  = {post_sim:.5f}")
assert abs(post_exact - post_sim) < 5e-3, "simulation should match Bayes"
print("Bayes' theorem verified numerically.")

Expected output (values approximate; the seed makes them reproducible):

n=      100: P_hat(A)=0.22000  |error|=0.05333
n=     1000: P_hat(A)=0.15800  |error|=0.00867
n=   100000: P_hat(A)=0.16632  |error|=0.00035
n=  1000000: P_hat(A)=0.16665  |error|=0.00002

P(D | +) analytic   = 0.16667
P(D | +) simulated  = 0.16621
Bayes' theorem verified numerically.

The Monte-Carlo loop is the classical skeleton of a quantum experiment: prepare, sample, count, estimate. The only thing Braket adds is that the per-shot probabilities come from $|\psi\rangle$ via the Born rule rather than from rng.integers.

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Exercises

1. (Easy) Complement and union. Given $\mathbb P(A) = 0.6$, $\mathbb P(B) = 0.5$, and $\mathbb P(A\cap B) = 0.3$, compute $\mathbb P(A^c)$, $\mathbb P(A\cup B)$, and $\mathbb P(A^c \cap B^c)$.

Solution

$\mathbb P(A^c) = 1 - 0.6 = 0.4$. By inclusion–exclusion $\mathbb P(A\cup B) = 0.6 + 0.5 - 0.3 = 0.8$. By De Morgan, $A^c\cap B^c = (A\cup B)^c$, so $\mathbb P(A^c\cap B^c) = 1 - 0.8 = 0.2$.

2. (Easy) Independence vs. disjointness. Suppose $\mathbb P(A) = 0.4$ and $\mathbb P(B)=0.5$. (a) If $A,B$ are independent, find $\mathbb P(A\cap B)$ and $\mathbb P(A\cup B)$. (b) If instead $A,B$ are disjoint, find the same quantities. (c) Can two events with positive probability be both independent and disjoint?

Solution

(a) Independent: $\mathbb P(A\cap B) = 0.4\cdot 0.5 = 0.2$, so $\mathbb P(A\cup B) = 0.4+0.5-0.2 = 0.7$. (b) Disjoint: $\mathbb P(A\cap B) = 0$, so $\mathbb P(A\cup B) = 0.9$. (c) No. Disjoint gives $\mathbb P(A\cap B)=0$, but independence requires $\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B) > 0$. The two are contradictory whenever both events have positive probability.

3. (Medium) Conditioning is a probability measure. Fix $B$ with $\mathbb P(B) > 0$. Prove that $\mathbb Q(A) := \mathbb P(A\mid B)$ satisfies the three Kolmogorov axioms on $(\Omega,\mathcal F)$.

Solution

(K1) $\mathbb Q(A) = \mathbb P(A\cap B)/\mathbb P(B) \ge 0$ since numerator $\ge 0$ (K1 for $\mathbb P$) and denominator $>0$. (K2) $\mathbb Q(\Omega) = \mathbb P(\Omega\cap B)/\mathbb P(B) = \mathbb P(B)/\mathbb P(B) = 1$. (K3) For pairwise disjoint $A_1,A_2,\dots$, the sets $A_i\cap B$ are also pairwise disjoint, and $(\bigcup_i A_i)\cap B = \bigcup_i (A_i\cap B)$. Hence

$$\mathbb Q\Big(\bigcup_i A_i\Big) = \frac{\mathbb P\big(\bigcup_i (A_i\cap B)\big)}{\mathbb P(B)} = \frac{\sum_i \mathbb P(A_i\cap B)}{\mathbb P(B)} = \sum_i \mathbb Q(A_i),$$

using countable additivity of $\mathbb P$. So $\mathbb Q$ is a probability measure. $\blacksquare$

4. (Medium) The Monty Hall problem. Three doors; a prize behind one. You pick door 1. The host, who knows where the prize is, opens a different door with no prize (door 3, say) and offers a switch to door 2. Using Bayes' theorem, compute $\mathbb P(\text{prize behind 2} \mid \text{host opened 3})$ and decide whether to switch.

Solution

Let $C_i$ = "prize behind door $i$", each with prior $\tfrac13$. Let $H_3$ = "host opens door 3" given you chose door 1. The likelihoods: $\mathbb P(H_3\mid C_1) = \tfrac12$ (prize behind your door; host picks 2 or 3 at random), $\mathbb P(H_3\mid C_2) = 1$ (prize behind 2; host must avoid it, opens 3), $\mathbb P(H_3\mid C_3) = 0$ (host never opens the prize door). Evidence: $\mathbb P(H_3) = \tfrac12\cdot\tfrac13 + 1\cdot\tfrac13 + 0 = \tfrac12$. Then

$$\mathbb P(C_2\mid H_3) = \frac{1\cdot\tfrac13}{\tfrac12} = \frac{2}{3}, \qquad \mathbb P(C_1\mid H_3) = \frac{\tfrac12\cdot\tfrac13}{\tfrac12} = \frac13 .$$

Switching wins with probability $\tfrac23$. Switch.

5. (Hard) Two-test sequential update. Reuse Example 1's disease ($\mathbb P(D)=0.01$, sensitivity $0.99$, false-positive rate $0.05$). You test positive twice with two conditionally independent tests (independent given true disease status). Compute the posterior $\mathbb P(D\mid +,+)$ two ways: (a) update once, then treat the posterior as the new prior and update again; (b) compute it in one shot from the joint likelihood. Confirm they agree.

Solution

(a) After one positive (Example 1), the posterior is $\pi_1 = \mathbb P(D\mid +) = 1/6 \approx 0.1667$. Use$\pi_1$ as the new prior. Conditional independence means the second test's likelihoods are unchanged ($0.99$ given $D$, $0.05$ given $D^c$):

$$\mathbb P(D\mid +,+) = \frac{0.99\,\pi_1}{0.99\,\pi_1 + 0.05(1-\pi_1)} = \frac{0.99\cdot\frac16}{0.99\cdot\frac16 + 0.05\cdot\frac56} = \frac{0.165}{0.165 + 0.041\overline{6}} \approx 0.7984 .$$

(b) One shot: likelihood of two positives is $0.99^2$ given $D$ and $0.05^2$ given $D^c$:

$$\mathbb P(D\mid +,+) = \frac{0.99^2(0.01)}{0.99^2(0.01) + 0.05^2(0.99)} = \frac{0.0098010}{0.0098010 + 0.0024750} \approx 0.7984 .$$

They agree. Sequential Bayesian updating is consistent with batch updating precisely when the observations are conditionally independent given the hypothesis — the mathematical reason we can accumulate evidence shot by shot.

6. (Hard) Boole's inequality (union bound). Prove that for any events $A_1, A_2, \dots$ (not necessarily disjoint), $\mathbb P\big(\bigcup_i A_i\big) \le \sum_i \mathbb P(A_i)$. Why is this bound ubiquitous in analyzing failure probabilities (e.g. "at least one of $m$ qubits errs")?

Solution

Disjointify: define $B_1 = A_1$ and $B_i = A_i \setminus \bigcup_{j<i} A_j$. The $B_i$ are pairwise disjoint, $B_i \subseteq A_i$, and $\bigcup_i B_i = \bigcup_i A_i$. By countable additivity (K3) and monotonicity,

$$\mathbb P\Big(\bigcup_i A_i\Big) = \mathbb P\Big(\bigcup_i B_i\Big) = \sum_i \mathbb P(B_i) \le \sum_i \mathbb P(A_i). \qquad\blacksquare$$

It is ubiquitous because it needs no independence assumption: to bound the chance that any of $m$ bad events happens (a qubit errs, a gate fails, a test gives a false positive), just sum the individual probabilities. If each of $m$ qubits errs with probability $\le p$, then $\mathbb P(\text{some qubit errs}) \le mp$ regardless of correlations — the basic budget behind error-correction thresholds.

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Checkpoint

1. State the three Kolmogorov axioms. Which one requires countable (not just finite) additivity, and why does that matter?
2. Why do we bother with a $\sigma$-algebra instead of just using all subsets of $\Omega$?
3. Give the definition of $\mathbb P(A\mid B)$ and explain in one sentence why it is itself a probability measure.
4. Two events are disjoint. Are they independent? Justify.
5. Write Bayes' theorem and name each of its four parts (prior, likelihood, posterior, evidence).
6. In Example 1, the test is $99\%$ sensitive yet a positive result yields only $\sim 17\%$ probability of disease. In one sentence, why?

Answers

1. (K1) $\mathbb P(A)\ge 0$; (K2) $\mathbb P(\Omega)=1$; (K3) countable additivity for disjoint events. (K3) is the countable one; countability is needed so we can take limits of events (e.g. tail events, convergence of sample frequencies in the LLN) and still get consistent probabilities.
2. For uncountable $\Omega$ (e.g. $[0,1]$ with uniform measure) it is impossible to assign a consistent probability to every subset (non-measurable sets exist, by Vitali). The $\sigma$-algebra names exactly the events we can measure; for finite/countable $\Omega$ we just take all subsets.
3. $\mathbb P(A\mid B) = \mathbb P(A\cap B)/\mathbb P(B)$ for $\mathbb P(B)>0$. It satisfies K1–K3 on $(\Omega,\mathcal F)$ (Exercise 3), so conditioning produces a legitimate probability space supported on $B$ — "shrink and renormalize."
4. Not (in general). If both have positive probability, disjointness gives $\mathbb P(A\cap B)=0 \ne \mathbb P(A)\mathbb P(B)$, so they are *dependent*. (The degenerate case where one has probability$0$ is the only exception.)
5. $\mathbb P(B_j\mid A) = \dfrac{\mathbb P(A\mid B_j)\mathbb P(B_j)}{\sum_i \mathbb P(A\mid B_i)\mathbb P(B_i)}$. $\mathbb P(B_j)$ = prior, $\mathbb P(A\mid B_j)$ = likelihood, $\mathbb P(B_j\mid A)$ = posterior, $\mathbb P(A) = \sum_i \mathbb P(A\mid B_i)\mathbb P(B_i)$ = evidence.
6. The disease is rare (prior $1\%$), so even a small false-positive rate applied to the large healthy population produces more false positives than the test produces true positives — the base rate dominates.

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Further Reading

• [CT, §2.1–2.5] Cover & Thomas — probability preliminaries underpinning information theory.
• [NC, §2.2.3–2.2.5] Nielsen & Chuang — measurement statistics; where these probabilities meet the Born rule.
• [Pre, Ch. 2] Preskill — the formalism, in the run-up to the quantum measurement postulate.
• Bill Bryson it isn't: for rigorous measure-theoretic probability, P. Billingsley, Probability and Measure (Wiley) is the standard graduate reference (the $\sigma$-algebra subtleties live here).

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