Question

If 2 percent of electric bulbs manufactured by a certain company are defective, Find the probability that is sample of 200 bulbs same 3 bulb are defectives.​

Answer 1

Answer:

cvvc

Step-by-step explanation:

Answer 2

Answer:

Step-by-step explanation

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