Question

It is known from past experience that in a certain factory 3% of products are defective. a sample of 100 items are taken at random. find the probability that exactly 5 products are defective.

Answer 1

Given:

In a certain factory 3% of products are defective.

A sample of 100 items are taken at random.

To find:

Find the probability that exactly 5 products are defective.

Solution:

From given, we have,

n = total number of items/samples = 100

p = Probability of defective = 3/100 = 0.03

x = No. of defective items  = 5

Now, we will find λ which is the parameter of Poisson distribution.

λ = n × p = 100 × 0.03 = 3

Now as we know,

$P(X) = \dfrac{\lambda^xe^{-\lambda}}{X!}$

P (X) = ( 3⁵ × e ^-3 ) / 5!

= (243 × 0.04978) / 120

= 0.1008

Therefore, the probability that exactly 5 products are defective is 0.1008.

Answer 2

Given :   in a certain factory 3% of products are defective. a sample of 100 items are taken at random

To find :  the probability that exactly 5 products are defective

Solution:

a certain factory 3% of products are defective

=> Probability of Defective  p = 3/100

Probability of non-defective q = 1 - 3/100  = 97/100

Sample of 100 items  => n = 100

Exactly 5 Defectives => x = 5

P(x) = ⁿCₓpˣqⁿ⁻ˣ

=> P(5) = ¹⁰⁰C₅(3/100)⁵(97/100)⁹⁵

= 0.1013

probability that exactly 5 products are defective  = 0.1013

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