1.

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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