23.2 Probability

Key Takeaways

  • Probability values live on [0,1]; the complement rule P(Aᶜ)=1−P(A) is the most-used identity on PA-CAT probability items
  • Mutually exclusive events use the addition rule P(A or B)=P(A)+P(B); independent events use the multiplication rule P(A and B)=P(A)·P(B)
  • Conditional probability P(A|B)=P(A and B)/P(B) re-scales the sample space to event B; independence is equivalent to P(A|B)=P(A)
  • Bayes' theorem inverts evidence and cause: P(disease|positive) depends on prevalence, sensitivity, and specificity together
  • Expected value E(X)=Σx·P(x) is the long-run average; for a clinical screening test it equals the weighted average payoff across all outcome branches
Last updated: August 2026

Probability as the Language of Uncertainty

On the PA-CAT, probability is the bridge between descriptive statistics and inference. The blueprint group Anticipating Patterns (Bulletin Table 11, PA-CAT Bulletin of Information, rev. 20240815) expects you to reason about chance events, conditional probabilities, and expected values in clinical-adjacent settings such as screening tests, adverse event counts, and patient flow. Every probability on the exam is a number between 0 and 1 inclusive, and the probabilities of all possible outcomes in a sample space sum to exactly 1.

The Four Fundamental Rules

RuleFormulaWhen It Applies
ComplementP(Aᶜ) = 1 − P(A)Always
Addition (general)P(A or B) = P(A) + P(B) − P(A and B)Any two events
Addition (mutually exclusive)P(A or B) = P(A) + P(B)A and B cannot co-occur
Multiplication (independent)P(A and B) = P(A)·P(B)A and B do not affect each other
Multiplication (general)P(A and B) = P(A)·P(BA)

The single most common PA-CAT error is using the simple addition rule when events overlap. If 30% of patients have hypertension and 20% have diabetes and 10% have both, P(HTN or DM) = 0.30 + 0.20 − 0.10 = 0.40, not 0.50. Always subtract the intersection.

Conditional Probability and Independence

Conditional probability re-scales the world to a sub-population: P(A|B) = P(A and B) / P(B). Read it as "the probability of A given that B has happened." Two events are independent when knowing B does not change the probability of A, i.e., P(A|B) = P(A). Independence is stronger than mutual exclusivity — mutually exclusive events are actually dependent, because if A occurred, B cannot.

A clinical example: if 5% of surgical patients develop an SSI and 8% are readmitted within 30 days, and the joint probability of SSI and readmission is 0.04, then P(readmission | SSI) = 0.04 / 0.05 = 0.80. Clearly readmission is far more likely given an SSI, so the two events are dependent. You could also verify independence by noting 0.80 ≠ 0.08.

Bayes' Theorem and the Screening Test Trap

Bayes' theorem inverts conditional probability to answer the clinically urgent question: given a positive test, what is the probability the patient actually has the disease? The formula:

P(D | +) = [P(+ | D) · P(D)] / [P(+ | D) · P(D) + P(+ | Dᶜ) · P(Dᶜ)]

where P(D) is prevalence, P(+|D) is sensitivity, and P(+|Dᶜ) is 1 − specificity (the false-positive rate). This is the positive predictive value (PPV).

Worked PA-CAT-Style Example

A clinic screens for a condition with prevalence 1%. The test has sensitivity 99% and specificity 95%. What is P(disease | positive)?

  • True positives: 0.99 × 0.01 = 0.0099
  • False positives: 0.05 × 0.99 = 0.0495
  • PPV = 0.0099 / (0.0099 + 0.0495) = 0.0099 / 0.0594 ≈ 0.167

So even with a seemingly excellent test, only about 1 in 6 positive screens is a true positive. The PA-CAT loves this counterintuitive result because it tests whether you remember that prevalence drives predictive value. The same test applied to a population with 20% prevalence yields PPV ≈ 0.83. Sensitivity and specificity are properties of the test; PPV and NPV are properties of the test in a population.

Random Variables and Expected Value

A random variable assigns a number to each outcome. A discrete random variable takes countable values; a continuous one takes values on an interval. The expected value E(X) = Σ x·P(x) is the long-run mean of the variable over many repetitions.

Worked Example: Number of No-Shows per Clinic Session

Suppose the number of no-shows X in a half-day clinic has the distribution below.

xP(x)x·P(x)
00.300.00
10.400.40
20.200.40
30.100.30

E(X) = 0 + 0.40 + 0.40 + 0.30 = 1.10 no-shows per session. Over 100 sessions you would expect about 110 no-shows. Expected value is a mean, not a guaranteed outcome — you cannot have 1.10 no-shows on a given day.

Variance of a Random Variable

The variance of a random variable is Var(X) = Σ (x − μ)² · P(x), the long-run average squared deviation. The standard deviation is σ = √Var(X). These population parameters (μ, σ) are distinct from the sample statistics (x̄, s) you met in Section 23.1; the PA-CAT often tests whether you can keep symbols straight.

Law of Total Probability

When outcomes partition the sample space (e.g., disease vs no disease), the law of total probability says P(B) = P(B|A)·P(A) + P(B|Aᶜ)·P(Aᶜ). The denominator of Bayes' theorem is just this law applied once. Recognizing it saves time: the PA-CAT will phrase items as "what is the overall probability of a positive test?" and you should compute P(+) = sensitivity·prevalence + (1−specificity)·(1−prevalence) before applying Bayes.

Common Probability Pitfalls on the PA-CAT

  1. Confusing P(A|B) with P(B|A). P(disease | positive) ≠ P(positive | disease). The first is PPV; the second is sensitivity. Bayes connects them.
  2. Assuming independence without justification. Two lab values from the same patient are usually dependent.
  3. Forgetting the complement. "At least one" problems are easiest via P(at least one) = 1 − P(none).
  4. Double-counting overlaps. Always subtract P(A and B) in the general addition rule.
  5. Treating expected value as the most likely value. E(X) need not equal any actual outcome.

Mastering these rules pays off across the remaining Statistics items, because sampling distributions and inference are built entirely on probability foundations.

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Test Your Knowledge

In a population with 1% disease prevalence, a test with 99% sensitivity and 95% specificity returns positive. What is the approximate probability the patient truly has the disease?

A
B
C
D
Test Your Knowledge

Which condition is equivalent to two events A and B being independent?

A
B
C
D
Test Your Knowledge

A clinic has E(X)=1.10 expected no-shows per session. Over 50 sessions, about how many total no-shows should be expected?

A
B
C
D