Question

Find the probability that at most 5 deffective defuses will be found in a box of 200.If experience shows that two percent of such defuses are defective.

Answer 1

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 = 2% = 1/50

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 * 1/50 = 4

Probability density as per Poisson's distribution function.  The probability that

the number of defective fuses is k is equal to:

P (x= k) = e^(-λ λ^k)k!

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

to k is equal to:

P ( x ≤ K ) ≤ e^(-λ(eλ^k))/k^k

Substitute λ = 4  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 4 defective fuses.  So there

eing only 5 or less defective fuses is very very small

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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^-4{ (4^0/0!) + (4^1/1!) + (4^2/2!) + (4^3/3!) + (4^4/4!) + (4^5/5!)}

= > 42.85 *e^-4