For such problems we have to assume that the samples of the fuses follow the standard normal probability distribution function or a poisson's probability function.
The number of defective fuses found in a sample of size N follows the Poisson's distribution function or standard normal distribution function.
p = 20% = 1/5
N = 200
Let X be the number of defective fuses in a box of N fuses.
Expected number of defective fuses in a box of 200 fuses is Np = λ = mean
E(X) = λ = 200 * 1/5 = 40
Probability density as per Poisson's distribution function. The probability that the number of defective fuses is k is equal to:
Also, the probability that the number of defective fuses is less than of equal to k is equal to:
Substitute λ = 40 and k = 5 in this expression :
P (X <= 5 ) <= 2.06 * 10⁻¹¹ very small probability.
It is expected as, in the box we expect around 40 defective fuses. So there being only 5 or less defective fuses is very very small.
You could also calculcate
P (X <= 5) = P(X=0) + P(X = 1)+P(X=2) + P(X=3) + P(X=4) +P(X=5)
If you assume the normal distribution function then:
μ = mean = 40