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If E and E2 are independent events, write the value of P((E1∪ E2)∩\((\bar{E}∪ \bar{E}2)\) |
Answer» 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 |
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