Question

The number of accidents is .4 accidents / day. What is the probability that the time from now to the next accident will be more than 3 days?What is the probability that the the time from now to the 3rd accident will be more than 5 days?

Answer 1

A) `P[X = 0 | X~"Poi"(lambda=1.2)] = 0.301 or 30.1%`
B) `P[X < 3 | X~"Poi"(lambda=2)] = 0.6767 or 67.67%`

Explanation:

Questions such as these which provide an "average" or mean number of events in a given duration of time are typically handled using distributions known as Poisson Distributions. There are some conditions we must check before using a Poisson Distribution:

• We must be given (or know) the mean number of events happening in a specific unit of time.
• We must be able to count the observed number of events in the same unit of time if we were to watch for them, and these counts must be from the natural number set `NN`.
• The likelihood of events happening within one specific unit of time must be independent of the likelihood of those kind of events happening in a different unit of time of the same duration.

In this problem, we are talking about accidents. We know that the mean number of accidents happens at a rate of .4 accidents/day. For our problem, we use a variable `lambda` (lambda) to store this value. Thus, `lambda = .4`.

Let us use the variable `X` to represent the number of accidents that will happen in one day. We will say that the probabilities of `X` happening are distributed according to a Poisson Distribution of parameter `lambda = .4`, or more commonly `X~"Poi"(lambda=.4)`.

According to the Poisson formulas, the probability of `x` events happening in the same length of time is given by the formula:

`P(X = x)=(lambda^x)/(x!e^lambda)`

For the first question, we want to know what the probability is that we will go more than 3 days without an accident. Since our original time duration was a single day (with .4 mean accidents happening during that time), we could adjust that to represent a Poisson distribution that has a duration of 3 days. This leads us to a new distribution `X~"Poi"(lambda=1.2)`, since .4 accidents/day leads to an average of (.4)(3) = 1.2 accidents/3 days.

Now we determine what the probability is of having 0 accidents in that Poisson distribution:

`P(X = 0) = lambda^0/(0!e^lambda) = 1.2^0/(e^1.2)=1/e^1.2~~0.301`

For the 2nd question we can proceed in a similar fashion. We're asked about the probability of going more than 5 days to get to the 3rd accident. We will derive a Poisson distribution of duration 5 days, or `X~"Poi"(lambda=2)`. (This is because (.4)(5) = 2).

In this distribution, `X` represents the number of accidents in 5 days, and since we'd like that to be less than 3, we seek `P(X<3)=sum_(x=0)^2P(X = x)`:

`=sum_(x=0)^2 lambda^x/(x!e^lambda) =2^0/(0!e^2)+2^1/(1!e^2)+2^2/(2!e^2)`
`color(white)(=sum_(x=0)^2 lambda^x/(x!e^lambda))~~0.1353+0.2707+0.2707`
`color(white)(=sum_(x=0)^2 lambda^x/(x!e^lambda))=0.6767`