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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:

P(X = k) = \frac{e^{-\lambda\ \lambda^k}}{k!}\\

Also, the probability that the number of defective fuses is less than of equal to k is equal to:

P(X \le k) \le \frac{e^{-\lambda\ (e\ \lambda)^k}}{k^k}\\

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)
=e^{-40}*[\frac{40^0}{0!} + \frac{40^1}{1!} + \frac{40^2}{2!} + \frac{40^3}{3!} + \frac{40^4}{4!} + \frac{40^5}{5!} ]\\\\=4.125 * 10^{-12}

If you assume the normal distribution function then:

   μ = mean = 40


3 5 3
could you please answer this
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