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 ?

Probability of a fuse lasts less than $0.8$ seconds is $0.8$. The probability of none of $n$ independent fuses last longer than $0.8$ seconds is therefore $0.8^n$. So the probability of at least one of them last longer than $0.8$ becomes $1-0.8^n$ which should be greater than or equal to $0.95$.

That is $$1-0.8^n\ge0.95,\,n\in\mathbb{N}$$

So $$n\ge\lceil\log(0.05)/\log(0.8)\rceil=14$$