Answer: Probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.
Step-by-step explanation:
Since we have given that
Mean rate defects per unit = 0.3
Since we will use "Poisson distribution":
[tex]P(X=K)=\frac{e^{-\lambda}\lambda^k}{k!}[/tex]
But we need to find the probability that a randomly selected unit will contain at least two surface-finish defect.
[tex]P(X\geq 2)=1+P(X=0)+P(X=1)[/tex]
So,
[tex]P(X=0)=\frac{e^{-0.3}0.3^0}{0!}=e^{-0.3}=0.74\\\\P(X=1)=\frac{e^{-0.3}\times 0.3}{1}=0.22[/tex]
so, it becomes,
[tex]P(X>2)=1-(0.74+0.22)=1-0.96=0.04[/tex]
Hence, probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.