Question

In one town, the number of burglaries in a week has a poisson distribution with a mean of 1.9. find the probability that in a randomly selected week the number of burglaries is at least three.

Answer 1

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) = $\frac{e^{-mean} mean^{x}}{x!}$

Using this formula probability of X=2,1,0 with mean = 1.9 is

P(X=2) = $\frac{e^{-1.9} 1.9^{2}}{2!}$

P(X=2) = $\frac{0.1495 * 3.61}{2}$

P(X=2) = 0.2698

P(X=1) = $\frac{e^{-1.9} 1.9^{1}}{1!}$

P(X=1) = $\frac{0.1495 * 1.9}{1}$

P(X=1) = 0.2841

P(X=0) = $\frac{e^{-1.9} 1.9^{0}}{0!}$

P(X=0) = $\frac{0.1495 * 1}{1}$

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