Question 841319
<pre>
Suppose we can expect some event to occur <font face="symbol">l</font> times over some interval of time.  
Then the probability that the event will occur exactly x times during a span
of that same length of time is given by this formula:

P(x;<font face="symbol">l</font>){{{""=""}}}{{{(lambda^x*e^(-lambda))/x!}}}

where {{{lambda}}}{{{""=""}}}{{{1.8}}} and {{{x}}}{{{""=""}}}{{{"0"}}}

Just substitute in the formula:

P(0;1.8){{{""=""}}}{{{(1.8^0*e^(-1.8))/0!}}}

Any number to the zero power is 1 and zero factorial is 1, so

P(0;1.8){{{""=""}}}{{{(1*e^(-1.8))/1}}}{{{""=""}}}{{{e^(-1.8)}}}{{{""=""}}}{{{0.1653}}}

Edwin</pre>