Question

Surface-finish defects in a small electric appliance occur at random with a mean rate of 0.3 defects per unit. find the probability that a randomly selected unit will contain at least two surface-finish defect.

Answer 1

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":

$P(X=K)=\frac{e^{-\lambda}\lambda^k}{k!}$

But we need to find the probability that a randomly selected unit will contain at least two surface-finish defect.

$P(X\geq 2)=1+P(X=0)+P(X=1)$

So,

$P(X=0)=\frac{e^{-0.3}0.3^0}{0!}=e^{-0.3}=0.74\\\\P(X=1)=\frac{e^{-0.3}\times 0.3}{1}=0.22$

so, it becomes,

$P(X>2)=1-(0.74+0.22)=1-0.96=0.04$

Hence, probability that a randomly selected unit will contain at least two surface- finish defect is 0.04.