# Find the probability that at most 5 defective fuses will be found in a box of 200, if experience shows that 20% of such fuses are defective.

1
by issh

2015-02-20T13:14:47+05:30

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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:

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.

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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)

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If you assume the normal distribution function then:

μ = mean = 40

P(