Question

If A1, A2, and A3 are three events and P( A1 ∩ A2) = P( A1 ∩ A3) 7= 0 but P( A2 ∩ A3) = 0, show that

P (at least one Ai ) = P( A1) + P( A2) + P( A3) − 2P( A1 ∩ A2).

Answer 1

By using the principle of inclusion-exclusion, the given probability conditions are applied to demonstrate the required equality that the probability of at least one event occurring is P(A1) + P(A2) + P(A3) - 2P(A1 ∩ A2).

Step-by-step explanation:

To show that the probability of at least one event occurring (P(at least one Ai)) equals P(A1) + P(A2) + P(A3) - 2P(A1 ∩ A2) when given P(A1 ∩ A2) = P(A1 ∩ A3) ≠ 0 and P(A2 ∩ A3) = 0, we'll use the principle of inclusion-exclusion for three events.

The probability of at least one event occurring is given by:

P(A1 ∨ A2 ∨ A3) = P(A1) + P(A2) + P(A3) - P(A1 ∩ A2) - P(A1 ∩ A3) - P(A2 ∩ A3) + P(A1 ∩ A2 ∩ A3)

Since P(A1 ∩ A2) = P(A1 ∩ A3) and P(A2 ∩ A3) = 0, we can simplify:

P(A1 ∨ A2 ∨ A3) = P(A1) + P(A2) + P(A3) - P(A1 ∩ A2) - P(A1 ∩ A2) - 0 + P(A1 ∩ A2 ∩ A3)

However, P(A1 ∩ A2 ∩ A3) is already included in both P(A1 ∩ A2) and P(A1 ∩ A3), so we subtract it twice. Thus, we get:

P(A1 ∨ A2 ∨ A3) = P(A1) + P(A2) + P(A3) - 2P(A1 ∩ A2)

Which proves the required equality.