Finding $n$ for the given probability. A certain explosive device will detonate if anyone of $n$ short-lived fuses lasts longer than $0.8$ seconds. Let $X_i$ represent the life of the $ith$ fuse. It can be assumed that each $X_i$ is uniformly distributed over the interval $(0,1)$ , also $X_i $'s are independent.
We need to find the number of fuses required if one wants to be 95% certain that the device will detonate.
That is : $P(detonation) = 0.95$
=> $P(X_1 > 0.8)+P(X_2 > 0.8)+P(X_3 > 0.8)$. . .     $+P(X_n > 0.8)=0.95$
=> $\sum_{i=1}^{n}P(X_i > 0.8)=0.95$.
Since $X_i$~$U(0,1)$ the above statement gives : $\sum_{i=1}^{n}(0.2)=0.95$
=> $0.2n = 0.95$ which gives $n \approx 5$ .
Is this correct ?

You are right that the probability of a detonation is $$P\left(\bigcup_{k=1}^n\{ X_k > 0.8\}\right)$$ but this is not just the sum because the events are not disjoint.
You have to subtract the intersections… 
