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

To find the probability of at least three burglaries in a week with an average of 1.9 burglaries, one must use the Poisson distribution. The cumulative probability of having 0 to 2 burglaries is subtracted from 1 to obtain the probability of three or more burglaries.

Step-by-step explanation:

The Poisson distribution is used for modeling the number of times an event occurs in a fixed interval of time or space. Given that the average number of burglaries in a week is 1.9, we need to find the probability of observing at least three burglaries in a given week. This can be expressed as P(X ≥ 3).

We calculate the cumulative probability of 0, 1, and 2 burglaries occurring and then subtract this value from 1 to find the probability of having at least 3 occurrences:

• Calculate P(X = 0), P(X = 1), and P(X = 2)
• Find P(X ≤ 2) by adding the probabilities of 0, 1, and 2 occurrences
• Subtract P(X ≤ 2) from 1 to get P(X ≥ 3)

This involves the use of the Poisson probability mass function and possibly a statistical calculator or software to compute the exact probabilities.

Answer 2

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) =
`(e^(-mean) mean^(x))/(x!)`

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

P(X=2) =
`(e^(-1.9) 1.9^(2))/(2!)`

P(X=2) =
`(0.1495 * 3.61)/(2)`

P(X=2) = 0.2698

P(X=1) =
`(e^(-1.9) 1.9^(1))/(1!)`

P(X=1) =
`(0.1495 * 1.9)/(1)`

P(X=1) = 0.2841

P(X=0) =
`(e^(-1.9) 1.9^(0))/(0!)`

P(X=0) =
`(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