Answer: Our required probability is 0.947.
Step-by-step explanation:
Since we have given that
Number of light bulbs selected = 30
Probability that the light bulb produced in a facility are defective = 2.8% = 0.028
We need to find the probability that fewer than 3 defective bulbs are found.
We will use "Binomial distribution":
n = 30, p = 0.028
so, P(X>3)=P(X=0)+P(X=1)+P(X=2)
So, it becomes,
[tex]P(X=0)=(1-0.0.28)^{30}=0.426[/tex]
and
[tex]P(X=1)=^{30}C_1(0.028)(0.972)^{29}=0.368\\\\P(X=2)=^{30}C_2(0.028)^2(0.972)^28=0.153[/tex]
So, the probability that fewer than three defective bulbs are defective is given by
[tex]0.426+0.368+0.153\\\\=0.947[/tex]
