Solution: $$ P(x=1)=P(x=2) $$ $$ \begin{array}{I} \frac{e^{-m} m^{1}}{1 !}=\frac{e^{-m} m^{2}}{2 !} \\ \therefore m=2 \end{array} $$ Thus, the mean & varience of poisson distribution be are both equal to m. $$ \begin{aligned} P(x=3)=& \frac{e^{-m} m^{3}}{3 !} \\ =& \frac{e^{-2}(2)^{3}}{3 !} \\ =& \frac{e^{-2} \times 8}{3 \times 2 \times 1} \\ =& \frac{4}{3 e^{2}} \\ =& 0.1805 \end{aligned} $$