Correct Answer - (i) `1/400` (ii) `19/200`
Let `E_(1)=` event of availability of the first engine.
And, `E_(2) =` event of availability of the second engine.
Then, `P(E_(1))=P(E_(2))=0.95` and `P(bar(E)_(1))=P(bar(E)_(2))=(1-0.95)=0.05`.
(i) P( neither of them is available when needed)
`=P (bar(E)_(1) and bar(E)_(2))=P(bar(E)_(1))xxP(bar(E)_(2))`.
(ii) P( an engine is available when needed)
`=P [(E_(1)" and not "E_(2)) or (E_(2)" and not "E_(1))]`
`=P[(E_(1) and bar(E)_(2)) or (E_(2) and bar(E)_(1))]`
`=P (E_(1) nn bar(E)_(2))+P(E_(2) nn bar(E)_(1))`
`={P (E_(1))xxP(bar(E)_(2))}+{P(E_(2))xxP(bar(E)_(1))}`.