The number of accidents that occur at a busy intersection is Poisson distributed with a mean of 4.5 per week. Find the probability of the following events.

A. No accidents occur in one week. Probability = 0.011111.

B. 5 or more accidents occur in a week.

C. One accident occurs today. Probability =

A. 0.1111

B. 0.1708 and 0

C. 0.338014.

Step-by-step explanation:

Requirement A:

A discrete random variable X is said to have a Poisson distribution if its probability function is given by,

f (x; λ) = (e^ (-λ) * λ^x) / x!; x = 0, 1, 2 ... ∞

where, e = 2.71828 and λ is the parameter of the distribution which is the mean number of success.

Given,

Mean number of accident λ = 4.5 per week

Let, X be the number of car accident which follow Poisson distribution. The probability of no accident occur in one week is,

P[no accident] = P[X=0]

= (e^ (-4.5) * 〖4.5〗^0) / 0!

= 0.1111

The probability of no accident is 0.1111.

Requirement B:

Given,

Mean number of accident λ = 4.5 per week

Let, X be the number of car accident which follow Poisson distribution. The probability of 5 accidents occurs in one week is,

P[exactly 5 accidents] = P[X=5]

= (e^ (-4.5) 〖4.5〗^5) / 5!

= 0.1708

The probability of no accident is 0.1708.

And, The probability of more than 5 accidents occurs in one week is,

P[more than 5 accidents] = P[X>5]

=1 - P[X is less than5]

= 1 - P[X=0] - P[X=1] - P[X=2] - P[X=3] - P[X=4] - P[X=5]

=1 - 0.1111 - 0.4999 - 0.11248 - 0.1687 - 0.1898 - 0.17083

= -.2521

= 0

The probability of more than 5 accidents is 0.

Requirement C:

Poisson distribution is related to the experimental results expressed in two ways: successful and failure under some limiting condition.

Given,

Mean number of accident λ = 4.5 per week

= 4.5/7 per day

=.6429

Let, X be the number of car accident which follow Poisson distribution. The probability of one accident occurs in one day is,

P[exactly 1 accident] = P[X=1]

= (e^ (-.6429) 〖.6429〗^1) / 1!

= 0.338015

The probability of no accident is 0.338015.