Answer:
(i). 0.03981
(ii).0.0048
Step-by-step explanation:
The probability density function of Poisson distribution is:
[tex]P(X=x,\lambda)=\frac{e^{-\lambda}\lambda^x}{x!} \; \;\;\,\; x=0,1,2,...[/tex]
Consider X is a number of typos error on a single page of a book and X follows the Poisson distribution with [tex]\lambda = \dfrac{1}{3}[/tex]
(i) Exactly two typos:
[tex]\begin{aligned}P(X = 2,\frac{1}{3})&=\frac{e^{-\frac{1}{3}}\frac{1}{3}^{2}}{2!}\\&=\frac{e^{-\frac{1}{3}}}{18}\\&=0.03981\end{aligned}[/tex]
(ii) Two or more typos:
[tex]\begin{aligned}P(X\geq2,\frac{1}{3})&=1-[P(X=0)+P(X=1)+P(X+2)]\\&=1-[0.7165+0.2388+0.03981]\\&=1-0.9952\\&=0.0048\end{aligned}[/tex]