Question

a sample of 300 bulbs is taken from a company.It is claimed by company that only 0.015 proportion of the bulbs is defective.what is the probability that a sample of 300 bulbs. any 4 bulbs are defective.atleast one is defective.almost 3 are defective​

Answer 1

Answer:

Since the probability of a bulb being defective is very small (p=0.02) and number of bulbs being moderately large, X (number of defective bulbs) may be assumed to be distributed as a Poisson distribution with parameter λ i.e. P(X=x)=e−λ(λ)xx! where λ=200∗.02=4

Now, P(X<2)⇒P(X=0)+P(X=1)

=e−λ+e−λλ

=e−λ(1+λ)

=5e−4=50.0183

Again, P(X>3)=1−P(X=0)−P(X=1)−P(X=2)

=1−e−λ(1+λ)−e−λλ22!

Calculation is left to you taking λ=4