A disease affects 1% of a population. A test is 90% sensitive (P(positive|disease) = 0.90) and 95% specific (P(negative|no disease) = 0.95). What is the overall probability that a randomly chosen person tests positive?
A90%, because the test is 90% accurate for people with the disease
B5.85%, by summing weighted conditional probabilities over the disease/no-disease partition
C1%, because only 1% of the population has the disease
D47.5%, by averaging the sensitivity and false-positive rate
The events {disease} and {no disease} partition the population. By the Law of Total Probability: P(positive) = P(positive|disease)·P(disease) + P(positive|no disease)·P(no disease) = 0.90·0.01 + 0.05·0.99 = 0.009 + 0.0495 = 0.0585. Option A is the most tempting wrong answer — it mistakes the conditional rate P(positive|disease) for the unconditional rate P(positive), ignoring that most of the population is disease-free and the false-positive rate dominates.
Question 2 Multiple Choice
You want to apply the Law of Total Probability to compute P(A). Which condition on your conditioning events B₁, B₂, B₃ is strictly required?
AThe events must be independent of A
BThe events must cover at least half the sample space
CThe events must be mutually exclusive and together cover the entire sample space
DEach event must have equal probability
The Law of Total Probability requires a partition: the Bᵢ must be mutually exclusive (no overlap — they can't both occur) AND exhaustive (they cover the whole sample space — one of them must occur). If the Bᵢ overlap, you double-count the probability of A in the overlap region. If they don't cover the whole space, you miss the contribution of A in the uncovered region. Equal probability (option D) is irrelevant — unequal partition pieces work fine. Independence from A (option A) is not required at all.
Question 3 True / False
If events B₁, B₂, …, Bₙ form a partition of the sample space, then the sum P(B₁) + P(B₂) + … + P(Bₙ) must equal 1.
TTrue
FFalse
Answer: True
This follows directly from the partition definition. The Bᵢ are mutually exclusive (no overlaps) and exhaustive (they cover the entire sample space). Since probability is additive over disjoint events and the Bᵢ cover everything, their probabilities must sum to exactly 1. This is also why the Law of Total Probability works: you are computing a weighted average of the conditional probabilities P(A|Bᵢ), with weights P(Bᵢ) that sum to 1, so the result is a valid probability.
Question 4 True / False
The Law of Total Probability applies to any collection of events B₁, …, Bₙ so long as they are mutually exclusive — it does not matter whether they cover the entire sample space.
TTrue
FFalse
Answer: False
Both conditions are required: mutual exclusivity AND exhaustiveness. If the events are mutually exclusive but don't cover the full sample space, then Σ P(A|Bᵢ)·P(Bᵢ) will undercount P(A) — it misses the probability contributed by outcomes outside all the Bᵢ. For example, if B₁ = {even numbers} and B₂ = {multiples of 3} in a roll of a die, these are not mutually exclusive (6 is in both) and don't exhaust the space. You must have a true partition to apply the law correctly.
Question 5 Short Answer
Why must the conditioning events form a partition (both mutually exclusive and exhaustive) for the Law of Total Probability to yield the correct answer? What goes wrong if each condition is violated separately?
Think about your answer, then reveal below.
Model answer: Mutual exclusivity ensures no double-counting: if two Bᵢ events can both occur, the probability of A ∩ Bᵢ ∩ Bⱼ would be added twice. Exhaustiveness ensures complete coverage: if some outcomes belong to no Bᵢ, the probability of A in that uncovered region is missed entirely. Together, the partition carves the sample space into non-overlapping strips that together account for all of probability — so summing the contributions P(A|Bᵢ)·P(Bᵢ) adds up to P(A) exactly once.
The geometric picture makes this clear: think of the sample space as a rectangle. A partition divides it into non-overlapping strips that together tile the whole rectangle. Event A is a blob that crosses these strips; P(A) equals the sum of A's area in each strip, which is exactly Σ P(A∩Bᵢ) = Σ P(A|Bᵢ)·P(Bᵢ). Overlapping strips would count some area of A twice; gaps would miss some area of A. Both conditions are load-bearing.