Correct Answer - 0.66976
Let `E_(1), E_(2), E_(3), E_(4)` be the respective events that the plane is hit in the 1st, 2nd, 3rd and 4 th shot. Then,
`P(E_(1))=0.4, P(E_(2))=0.3, P(E_(3))=0.2` and `P(E_(4))=0.1`
`:. P(bar(E)_(1))=(1-0.4)=0.6, P (bar(E)_(2))=(1-0.3)=0.7`,
`P(bar(E)_(3))=(1-0.2)=0.8, P(bar(E)_(4))=(1-0.1)=0.9`.
`:.` P(at least one shot hits the plane) `=1-P (bar(E)_(1) and bar(E)_(2) and bar(E)_(3) and bar(E)_(4))`
`=1-{P(bar(E)_(1))xxP(bar(E)_(2))xxP(bar(E)_(3))xxP(bar(E)_(4))}`.