The Probability that at least one of the events `E_(1)` and `E_(2)` will occur is 0.6. If the probability of their occurrence simultaneously is 0.2, then find `P(barE_(1))+P(barE_(2))`

Answer

N/a

Explanation

Given that `P(E_(1)uuE_(2))=0.6` and `P(E_(1)nnE_(2))=0.2`
Now `P(E_(1)uuE_(2))=P(E_(1))+P(E_(2))-P(E_(1)nnE_(2))`
`0.6=P(E_(1))+P(E_(2))-0.2`
`implies 0.8=1-P(barE_(1))+1-P(barE_(2))`
`implies P(barE_(1))+P(barE_(2))=2-0.8`
`=1.2.`