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
$m=np=10\times0.2=2$
$\therefore $ The probability that not more than one defective is found.
$=P\left(k=0\right)+P\left(k=1\right)-e^{-m}+me^{-m}=e^{-2}+2e^{-2}=3e^{-2}$