Question 460443
If the expected number of occurrences in this interval is a, then theprobability that there are exactly k occurrences is equal to
{{{P(k,a )=((a^k)e^(-a))/k!}}}, where
e is the base of the natural logarithm (e = 2.71828...)
k is the number of occurrences of an event — the probability of which is given by the function
k! is the factorial of k
a is a positive real number, equal to the expected number of occurrences during the given interval. 

Babies arrive at an average of 8 babies per hour, so {{{a=8}}}
1) what is the probability exactly 7 babies are born during the next hour?
{{{k=7}}}
{{{P(7,8 )=((8^7)e^(-8))/7!=0.1396}}}
2) what is the probability that fewer than 3 babies are born during the next hour?
Fewer than 3 - means that 0 babies (k=0) or 1 baby (k=1) or 2 babies (k=2) are born during the next hour
{{{P(fewer than 3)=P(0,8)+P(1,8)+P(2,8)}}}
{{{P(fewer than 3)=((8^0)e^(-8))/0!+((8^1)e^(-8))/1!+((8^2)e^(-8))/2!=0.0003+0.0027+0.0107=0.0137}}}