The probability of a Poisson distribution is given by:
[tex]P(k)=e^{-\lambda} \frac{\lambda^k}{k!} [/tex]
where: [tex]\lambda[/tex] is the average number of events and k is the required probability.
Given that a manufacturer packages bolts in boxes containing 100 each and that each box of 100 bolts contains on average four defective bolts. Thus, [tex]\lambda=4[/tex].
The probability that the box will contain less than three defective bolts is given by:
[tex]P(0)+P(1)+P(2)=e^{-4}\cdot \frac{4^0}{0!} +e^{-4}\cdot \frac{4^1}{1!} +e^{-4}\cdot \frac{4^2}{2!} \\ \\ =e^{-4}(1+4+8)=13(0.0183)=0.2381[/tex]
