Question

If electricity power failures occur according to a Poisson distribution with an average of 3 failures every twenty weeks, calculate the probability that there will not be more than one failure during a particular week.

Answer 1

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.

Explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

In which

x is the number of sucesses

e = 2.71828 is the Euler number

`\mu` is the mean in the given interval.

3 failures every twenty weeks

This means that for 1 week,
`\mu = (3)/(20) = 0.15`

Calculate the probability that there will not be more than one failure during a particular week.

Probability of at most one failure, so:

`P(X \leq 1) = P(X = 0) + P(X = 1)`

Then

`P(X = x) = (e^(-\mu)*\mu^(x))/((x)!)`

`P(X = 0) = (e^(-0.15)*0.15^(0))/((0)!) = 0.8607`

`P(X = 1) = (e^(-0.15)*0.15^(1))/((1)!) = 0.1291`

Then

`P(X \leq 1) = P(X = 0) + P(X = 1) = 0.8607 + 0.1291 = 0.9898`

0.9898 = 98.98% probability that there will not be more than one failure during a particular week.