Question 327628
The Poisson is given by 

{{{P(x)=L^x*e^(-L)/x!}}}

In this case, {{{L=15000*(.00006)=.9}}}

So, AT MOST 2 means 2 and less. In other words, x=0,1,2.

So, we have to add up the three cases for 0,1,and 2.

{{{(.9)^0*e^(-.9)/0!+(.9)^1*e^(-.9)/1!+(.9)^2*e^(-.9)/2!=.937}}}