Question 1184023
The pmf for the Poisson distribution is {{{p(x) = e^(-lambda)*(lambda^(x)/x!)}}}  for x = 0, 1, 2, 3, 4, ...


i.  {{{P(X >= 3) = 1 - p(0) - p(1) - p(2) = 1 - e^(-3.1)*(3.1^(0)/0!) - e^(-3.1)*(3.1^(1)/1!) - e^(-3.1)*(3.1^(2)/2!) = 0.59884}}}, to 5 d.p.


ii.    {{{p(0) = e^(-3.1)*(3.1^(0)/0!) = 0.04505}}}, to 5 d.p.