Question

The odds that some machine malfunctioned in its first three years of operation were 0.003. If 40 machines are randomly chosen, calculate the probability that 38 or more will not fail in the first three years.

Answer 1

Let $X=$ the number of machines which will fail in the first three years: $X\sim Bin(n, p)$

$$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$

Given that $\dfrac{p}{1-p}=0.003, n=40$

Then $p=\dfrac{3}{1003}\approx0.003$

$$P(X\leq2)=P(X=0)+P(X=1)+P(X=2)=$$

$$=\binom{40}{0}(0.003)^0(1-0.003)^{40-0}+$$

$$+\binom{40}{1}(0.003)^1(1-0.003)^{40-1}+$$

$$+\binom{40}{2}(0.003)^2(1-0.003)^{40-2}=$$ $$=0.8867605+0.1067315+0.0062626\approx0.999755$$

$$\approx0.993511$$

When the value of $n$ in a binomial distribution is large and the value of $p$  is very small, the binomial distribution can be approximated by a Poisson distribution. If $n>20$ and $np<5$  or $n(1-p)<5$ then the Poisson is a good approximation.

We have $n=40>20, np=40(0.003)=0.12<5.$

Use the Poisson Approximation to Binomial Distribution

$$\lambda=np=40(0.003)=0.12$$

$$P(X=x)={e^{-\lambda}\lambda^x\over x!}$$

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

$$={e^{-0.12}(0.12)^0\over 0!}+{e^{-0.12}(0.12)^1\over 1!}+{e^{-0.12}(0.12)^2\over 2!}=$$

$$=e^{-0.12}(1+0.12+0.0072)\approx0.999737$$