Question

Suppose that the probability that an item produced by a particular machine is defective equals 0.2. If 10 items produced from this machine are selected at random, the probability that not more than one defective is found

• A. $\frac{1}{e^2}$
• B. $\frac{2}{e^2}$
• C. $\frac{3}{e^2}$
• D. none of these

Answer 1

Answer: (C)

We suppose the distribution to be Binomial with $n=10,p=0.2,q=1-p=0.8$
$\therefore$ The probability that not more than one defective is found.
$=P\left(k=0\right)+P\left(k=1\right)=q^n+\left(\frac{n}{1}\right)p{q}^{n-1}$
$={\left(0.8\right)}^{10}+10\left(0.2\right){\left(0.8\right)}^9={\left(0.8\right)}^9=\left[0.8+2\right]=2.8{\left(0.8\right)}^9$.
This value is very small so the Binomial probabilities areapproximated by Poisson probabilities then
$m=np=10\times 0.2=2$
$\therefore$ The probability that not more than one defective is found.
$=P\left(k=0\right)+P\left(k=1\right)-e^{-m}+m{e}^{-m}=e^{-2}+2e^{-2}=3e^{-2}$