Question

On Texas Avenue between University Drive and George Bush Drive, accidents occur according to a Poisson process at a rate of three accidents per week. (a) What is the probability that a given day has no accidents? (b) Whenever there is an accident, the risk is 1 in 8 that personal injury will result. What is the probability that February (which has four weeks) has at least one accident with a personal injury?

Answer 1

(a) The probability is 0.6514

(b) The probability is 0.7769

Explanation:

If the number of accidents occur according to a poisson process, the probability that x accidents occurs on a given day is:

`P(x)=(e^(-at)*(at)^(x) )/(x!)`

Where a is the mean number of accidents per day and t is the number of days.

So, for part (a), a is equal to 3/7 and t is equal to 1 day, because there is a rate of 3 accidents every 7 days.

Then, the probability that a given day has no accidents is calculated as:

`P(x)=(e^(-3/7)*(3/7)^(x))/(x!)`

`P(0)=(e^(-3/7)*(3/7)^(0))/(0!)=0.6514`

On the other hand the probability that February has at least one accident with a personal injury is calculated as:

P(x≥1)=1 - P(0)

Where P(0) is calculated as:

`P(x)=(e^(-at)*(at)^(x) )/(x!)`

Where a is equivalent to (3/7)(1/8) because that is the mean number of accidents with personal injury per day, and t is equal to 28 because 4 weeks has 28 days, so:

`P(x)=(e^(-(3/7)(1/8)(28))*((3/7)(1/8)(28))^(x))/(x!)`

`P(0)=(e^(-(3/7)(1/8)(28))*((3/7)(1/8)(28))^(0))/(0!)=0.2231`

Finally, P(x≥1) is:

P(x≥1) = 1 - 0.2231 = 0.7769