Question

Show that the additive law of probability holds for conditional probabilities. That is, if A, B, and C are events such that P(C) > 0, prove that P(A ∪ B|C) = P(A|C) + P(B|C) − P(A ∩ B|C)

Answer 1

The additive law of probability holds for conditional probabilities. P(A ∪ B|C) = P(A|C) + P(B|C) - P(A ∩ B|C).

Step-by-step explanation:

To show that the additive law of probability holds for conditional probabilities, we need to prove that P(A ∪ B|C) = P(A|C) + P(B|C) - P(A ∩ B|C). Let's break down the proof step by step:

Start with the definition of conditional probability: P(A|C) = P(A ∩ C) / P(C)

Similarly, P(B|C) = P(B ∩ C) / P(C)

Now, let's focus on the left-hand side of the equation: P(A ∪ B|C)

Using the definition of conditional probability, P(A ∪ B|C) = P((A ∪ B) ∩ C) / P(C)

Applying the distributive property, P((A ∪ B) ∩ C) = (P(A ∩ C) + P(B ∩ C)) - P((A ∩ B) ∩ C)

We can now substitute the expressions from step 1 and step 2 into step 5: P((A ∪ B) ∩ C) = (P(A|C) * P(C) + P(B|C) * P(C)) - P((A ∩ B) ∩ C)

Cancelling out P(C) on both sides, we get: P(A ∪ B|C) = P(A|C) + P(B|C) - P((A ∩ B) ∩ C) / P(C)

Finally, we notice that P((A ∩ B) ∩ C) / P(C) = P(A ∩ B|C), so we can substitute this back into the equation: P(A ∪ B|C) = P(A|C) + P(B|C) - P(A ∩ B|C)