Question

The number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a mean of 0.025 failure per hour.

a. What is the probability that the instrument does not fail in an 8-hour shift?
b. What is the probability of at least one failure in a 24-hour day?

Answer 1

a) There is an 81.87% probability that the instrument does not fail in an 8-hour shift.

b) There is a 45.12% probability of at least one failure in a 24-hour day.

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 time interval.

a. What is the probability that the instrument does not fail in an 8-hour shift?

The mean for an hour is 0.025 failures.

For 8 hours, we have
`\mu = 8*0.025 = 0.2`

This probability is P(X = 0).

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

`P(X = 0) = (e^(-0.2)*(0.2)^(0))/((0)!) = 0.8187`

There is an 81.87% probability that the instrument does not fail in an 8-hour shift.

b. What is the probability of at least one failure in a 24-hour day?

The mean for an hour is 0.025 failures.

For 24 hours, we have
`\mu = 24*0.025 = 0.6`

Either we have at least one failure, or we have no failures. The sum of the probabilities of these events is decimal 1. So

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

`P(X \geq 1) = 1 - P(X = 0)`

In which

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

`P(X = 0) = (e^(-0.6)*(0.2)^(0))/((0)!) = 0.5488`

So

`P(X \geq 1) = 1 - P(X = 0) = 1 - 0.5488 = 0.4512`

There is a 45.12% probability of at least one failure in a 24-hour day.