Probability that a bomb dropped from …

probability that a bomb dropped from a plane hits a target is 0.4
e.g., P(E) = 0.4 = 2/5
so, probability that a bomb didn't drop from a plane is = 1 - 0.4 = 0.6
so, P(E') = 0.6 = 3/5

now, we have to find out, probability that the bridge will be distroyed.
here, question mentioned that, minimum two bombs are required for distroying the bridge. so, we have to find P(x ≥ 2)
e.g., P(x ≥ 2) = 1 -{ P(x = 0) + P(x = 0)}

according to binomial probability concept
     P(x ) = ⁿCₓ{P(E)ˣ.P(E')ⁿ⁻ˣ
here, n indicates the total number of bombs ,x indicates how much bombs we used.
  now, P(x = 0) = ⁶C₀{2/5}⁰.{3/5}⁶⁻⁰ = {3/5}⁶
P(x = 1) = ⁶C₁{2/5}¹{3/5}⁶⁻¹ = ⁶C₁(2/5)(3/5)⁵
 
so, P(x ≥ 2) = 1 - [(3/5)⁶ + ⁶C₁(2/5)(3/5)⁵ ]
                   = 1 - [ 729/15625 + 2912/15625] = 1 - 3641/15625
                   = 11984/15625 = 0.767