Answer
8 have 0 defects $0.094$
3 have 1 defect $0.030$
1 has 2 defects $0.297$
Work Step by Step
First we calculate the mean values:
0 defects $\mu0=np=12\times0.9=10.8$
1 defects $\mu1=12\times0.06=0.72$
2 defects $\mu2=12\times0.04=0.48$
Use the Poisson formula: $P(x)=\frac{\mu^xe^{-\mu}}{x!}$
8 have 0 defects $P_0(8)=\frac{10.8^8e^{-10.8}}{8!}=0.094$
3 have 1 defect $P_1(3)=\frac{0.72^3e^{-0.72}}{3!}=0.030$
1 has 2 defects $P_2(1)=\frac{0.48^1e^{-0.48}}{1!}=0.297$
