Conditional Probability and Discrete Distributions

Core Probability Definitions and Formulas

• Independence: Events A and B are independent if P(A ∩ B) = P(A)P(B).
• Mutual Exclusivity: Events A and B are mutually exclusive if P(A ∩ B) = 0.

Key Relationships

• If P(A) + P(B) > 1, then A and B are not mutually exclusive. (True)
• If P(A) + P(B) = 1, A and B are not necessarily mutually exclusive. (False, unless A and B are complements)

Conditional Probability and Addition Rule

• Multiplication Rule: P(A ∩ B) = P(A) P(B|A).
• Conditional Probability Definition: P(B|A) = P(A ∩ B) / P(A), provided P(A) > 0.
• Addition Rule: P(A ∪ B) = P(A) + P(B) - P(A ∩ B).

Advanced Probability Relationships

Bayes' Theorem Setup

Given P(A|B) = p, P(B|A) = q, and P(A) = r. Using the definition of conditional probability:

P(A|B) = [P(B|A) × P(A)] / P(B)

Substituting the variables: p = (q × r) / P(B).

Therefore, P(B) = (q × r) / p.

Independence and Mutual Exclusivity

If A and B are independent, they are not necessarily mutually exclusive (unless P(A)=0 or P(B)=0). If P(A) > 0 and P(B) > 0, independence implies P(A ∩ B) > 0, meaning they are not mutually exclusive.

Subset Condition Check

P(A ∩ B) ≥ P(B)

This inequality implies that B must be a subset of A (B ⊆ A). If B ⊆ A, then P(A ∩ B) = P(B).

Case Study: Discrete Probability Distribution X

1. Sample Space Definition

The sample space S for the random variable X is:

S = {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 18, 20, 24, 25, 30, 36}

2. Probability Mass Function (PMF)

Probabilities are defined based on the number of rolls required:

• One Roll: P(X=1) = P(X=2) = P(X=3) = 1/6.
• Two Rolls (Multiple of 4): If k is a multiple of 4 (but not 12, 24, or 36), P(X=k) = (1/6)(1/4) = 1/24.
• Two Rolls (Multiple of 5): If k is a multiple of 5, P(X=k) = (1/6)(1/5) = 1/30.
• Two Rolls (Multiple of 6): If k is a multiple of 6, P(X=6) = (1/6)(1/6) = 1/36.

3. Calculating Specific Probabilities

• P(X=12) = P(4, 3) + P(6, 2) = 1/24 + 1/36.
• P(X=15) = P(5, 3) = 1/30.

Conditional Probability Example: P(X=15 | 1st roll = 5)

P(1st roll = 5) = 1/6.

Since X=15 requires the first roll to be 5, X=15 is a subset of the event "1st roll = 5".

P(X=15 | 1st roll = 5) = P(X=15) / P(1st roll = 5) = (1/30) / (1/6) = 1/5.

4. Dependence Conclusion

The event (X ≤ 6) is a subset of (X ≤ 12). Since one event is contained within the other, they are dependent (unless the probability of the subset is zero or the superset is the entire sample space).

The event (X is even) and the entire sample space S (X < 40) are independent because any event A is independent of the entire sample space S: P(A | S) = P(A).

Application: Medical Test Accuracy

Context and Assumptions

Let P be the event that a woman is pregnant, and let $p = P(P)$ be the prevalence rate. We assume the test has:

• Sensitivity (True Positive Rate): 0.94 (1 - 0.06 False Negative Rate).
• Specificity (True Negative Rate): 0.95 (1 - 0.05 False Positive Rate).

1. Multiple Independent Tests (False Negative Scenario)

If a pregnant woman takes two independent tests, the probability that both tests yield an inaccurate result (false negative) is:

P(Test 1 Inaccurate ∩ Test 2 Inaccurate | P) = (0.06) × (0.06) = 0.0036.

2. Multiple Independent Tests (False Positive Scenario)

If a non-pregnant woman takes two independent tests, the probability that both tests yield an inaccurate result (false positive) is:

P(Test 1 Inaccurate ∩ Test 2 Inaccurate | P′) = (0.05) × (0.05) = 0.0025.

3. Calculating Overall Accuracy

Using the Law of Total Probability, the overall probability of an accurate result (A) is:

P(Accurate) = P(Accurate | P)P(P) + P(Accurate | P′)P(P′)

P(Accurate) = p(0.94) + (1 - p)(0.95)

P(Accurate) = 0.94p + 0.95 - 0.95p

P(Accurate) = 0.95 - 0.01p

Final Note on Concepts

Mutually Exclusive: Events where only one outcome can occur (e.g., flipping a coin results in heads OR tails).

Independent: Events where the outcome of one does not depend on the other.