Probability axioms and theorems — addition, multiplication, Bayes' theorem when to use

Question

A bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Find $P(\text{second is red} | \text{first is blue})$ . Also, if a test for a disease has 99% sensitivity and 95% specificity, and 1% of the population has the disease, find the probability that a person who tests positive actually has the disease (using Bayes’ theorem).

Solution — Step by Step

Given: first ball is blue. After drawing one blue ball, remaining = 5 red + 2 blue = 7 balls.

P(\text{2nd red} | \text{1st blue}) = \frac{5}{7} \approx \mathbf{0.714}

This is a direct application of the conditional probability definition.

Let $D$ = has disease, $T^+$ = tests positive.

Given: $P(D) = 0.01$ , $P(T^+ | D) = 0.99$ (sensitivity), $P(T^- | D') = 0.95$ so $P(T^+ | D') = 0.05$ (false positive rate).

P(D | T^+) = \frac{P(T^+ | D) \cdot P(D)}{P(T^+ | D) \cdot P(D) + P(T^+ | D') \cdot P(D')}

P(D | T^+) = \frac{0.99 \times 0.01}{0.99 \times 0.01 + 0.05 \times 0.99} = \frac{0.0099}{0.0099 + 0.0495} = \frac{0.0099}{0.0594} \approx \mathbf{0.167}

Only about 16.7% chance the person actually has the disease despite testing positive. This counterintuitive result shows why Bayes’ theorem matters — the low prevalence (1%) means false positives dominate.

Why This Works

graph TD
    A["Probability: Which theorem?"] --> B["Events A OR B?"]
    B --> C["Addition: P A∪B = P A + P B - P A∩B"]
    A --> D["Events A AND B?"]
    D --> E["Independent? P A∩B = P A × P B"]
    D --> F["Dependent? P A∩B = P A × P B|A"]
    A --> G["Reverse conditional?"]
    G --> H["Bayes: P A|B = P B|A × P A / P B"]
    A --> I["Sequence of events?"]
    I --> J["Total probability: P B = Σ P B|Aᵢ P Aᵢ"]

Addition theorem: use when asking “what is the probability of A OR B happening?”
Multiplication theorem: use when asking “what is the probability of A AND B both happening?”
Bayes’ theorem: use when you know $P(B|A)$ but need $P(A|B)$ — reversing the condition.
Total probability: use when event B can happen through multiple mutually exclusive paths $A_1, A_2, ...$

Alternative Method

For the Bayes’ theorem problem, a tree diagram or frequency table approach is more intuitive:

Imagine 10,000 people. 100 have the disease (1%). Of these 100, 99 test positive (99% sensitivity). Of the 9,900 healthy, 495 test positive (5% false positive).

Total positive = $99 + 495 = 594$ . Of these, $99$ actually have the disease.

$P(D | T^+) = 99/594 \approx 16.7\%$ . Same answer, but the reasoning is concrete and less error-prone.

Common Mistake

Confusing $P(A|B)$ with $P(B|A)$ . These are NOT the same. $P(\text{positive} | \text{disease})$ = 99% (sensitivity), but $P(\text{disease} | \text{positive})$ = 16.7% (the question we actually care about). This confusion is called the “prosecutor’s fallacy” and appears in JEE and CBSE problems regularly. Whenever a problem gives you one conditional probability and asks for the reverse, it is a Bayes’ theorem problem.

Addition: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Multiplication: $P(A \cap B) = P(A) \cdot P(B|A)$

Bayes’ theorem: $P(A_i|B) = \frac{P(B|A_i) \cdot P(A_i)}{\sum_j P(B|A_j) \cdot P(A_j)}$

Total probability: $P(B) = \sum_{i} P(B|A_i) \cdot P(A_i)$

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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