Show that E(F ∪ G) = EF ∪ EG.
E(F ∪ G) is the expected value of the union of two events F and G. The expected value of an event is calculated as the sum of the product of each possible outcome and its corresponding probability. EF = Σ(x * P(x)) for x in F EG = Σ(x * P(x)) for x in G E(F ∪ G) = Σ(x * P(x)) for x in (F ∪ G) = Σ(x * P(x)) for x in F + Σ(x * P(x)) for x in G - Σ(x * P(x)) for x in (F ∩ G) As we know that EF = Σ(x * P(x)) for x in F and EG = Σ(x * P(x)) for x in G So, E(F ∪ G) = EF + EG - Σ(x * P(x)) for x in (F ∩ G) Now, we can see that E(F ∪ G) = EF + EG, as the expected value of the union of two events F and G is equal to the sum of the expected values of each individual event . It's important to note that this equation holds only if F and G are mutually exclusive events.
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