🔬 Session 2.4 - Unions of Events and Addition Rules

We have data from a semiconductor process. There are 940 wafers total. Each wafer is classified by contamination level and location in the sputtering tool.

Let:

• H: wafer has high contamination → 358/940
• C: wafer came from center → 626/940
• H ∩ C: wafer has high contamination and came from center → 112/940
• H ∪ C: wafer is from center or has high contamination (or both)

We use the addition rule:

P(H ∪ C) = P(H) + P(C) − P(H ∩ C)

P(H ∪ C) = 358/940 + 626/940 − 112/940 = 872/940 ≈ 0.9277

Interpretation: Most wafers are either from the center, have high contamination, or both.

We use pH levels as an example. Let X be the pH value of a water sample.

We want to find the probability that X is between 6.5 and 7.8. We can divide that range into mutually exclusive parts — no overlap.

Let:

• P(6.5 ≤ X ≤ 7.0) = 0.30
• P(7.0 < X ≤ 7.5) = 0.45
• P(7.5 < X ≤ 7.8) = 0.15

Total Probability: P(6.5 < X ≤ 7.8) = 0.30 + 0.45 + 0.15 = 0.90

This works because the events are mutually exclusive (they do not overlap).

We are given the following probabilities:

• P(A) = 0.5
• P(B) = 0.6
• P(C) = 0.4
• P(A ∩ B) = 0.3
• P(A ∩ C) = 0.2
• P(B ∩ C) = 0.25
• P(A ∩ B ∩ C) = 0.1

Goal: Calculate P(A ∪ B ∪ C)

Formula:

P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C)

Substituting:

P(A ∪ B ∪ C) = 0.5 + 0.6 + 0.4 − 0.3 − 0.2 − 0.25 + 0.1 = 0.85

Interpretation: The total probability that a case is in A, B, or C is 0.85 (some overlap exists).