A radioactive object emits particles according to a Poisson

A radioactive object emits particles according to a Poisson process at an
average rate of 7.5 particles per second. We observe the object for a total of 5 seconds.
Suppose it is known that at least 20 particles will be emitted during this interval, what is
the probability that no more than 25 particles will be emitted during this interval?

Solution

Possion Distribution
PMF of P.D is = f ( k ) = e- x / x!
Where   
= parameter of the distribution.
x = is the number of independent trials
average rate of 7.5 particles per second
average rate in 5 seconds = 5*7.5 = 37.5
P( X = 20 ) = e ^-37.5 * 37.5^20 / 20! = 0.0006
P( X = 21 ) = e ^-37.5 * 37.5^21 / 21! = 0.0011
P( X = 22 ) = e ^-37.5 * 37.5^22 / 22! = 0.002
P( X = 23 ) = e ^-37.5 * 37.5^23 / 23! = 0.0032
P( X = 24 ) = e ^-37.5 * 37.5^24 / 24! = 0.005
P( X = 25 ) = e ^-37.5 * 37.5^25 / 25! = 0.0075

P( 20 < = X < = 25) = P(X=25) + P(X=24) + P(X=23) + P(X=22) + P(X=21) + P(X=20) = 0.0075 + 0.005 + 0.0032 + 0.002 + 0.0011 + 0.0006 = 0.0194