Let X be the number of burglaries in a week. X follows Poisson distribution with mean of 1.9
We have to find the probability that in a randomly selected week the number of burglaries is at least three.
P(X ≥ 3 ) = P(X =3) + P(X=4) + P(X=5) + ........
= 1 - P(X < 3)
= 1 - [ P(X=2) + P(X=1) + P(X=0)]
The Poisson probability at X=k is given by
P(X=k) = [tex] \frac{e^{-mean} mean^{x}}{x!} [/tex]
Using this formula probability of X=2,1,0 with mean = 1.9 is
P(X=2) = [tex] \frac{e^{-1.9} 1.9^{2}}{2!} [/tex]
P(X=2) = [tex] \frac{0.1495 * 3.61}{2} [/tex]
P(X=2) = 0.2698
P(X=1) = [tex] \frac{e^{-1.9} 1.9^{1}}{1!} [/tex]
P(X=1) = [tex] \frac{0.1495 * 1.9}{1} [/tex]
P(X=1) = 0.2841
P(X=0) = [tex] \frac{e^{-1.9} 1.9^{0}}{0!} [/tex]
P(X=0) = [tex] \frac{0.1495 * 1}{1} [/tex]
P(X=0) = 0.1495
The probability that at least three will become
P(X ≥ 3 ) = 1 - [ P(X=2) + P(X=1) + P(X=0)]
= 1 - [0.2698 + 0.2841 + 0.1495]
= 1 - 0.7034
P(X ≥ 3 ) = 0.2966
The probability that in a randomly selected week the number of burglaries is at least three is 0.2966
