As E1 and E2 are independent events.
∴ P((E1∪ E2)∩(E1’ ∩ E2’))
As by De Morgan’s law.
E1’ ∩ E2’ = (E1∪ E2)’
∴ P((E1∪ E2)∩(E1’ ∩ E2’)) = P((E1∪ E2)∩(E1∪ E2)’)
⇒ P((E1∪ E2)∩(E1∪ E2)’) = P(empty set) = 0
Couldn't generate an explanation.
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