The number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a mean of 0.030 failures 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 1 failure in a 24-hour day? Round your answers to four decimal places (e. g. 98.7654).

a) The probability of not having failures in 8 days is 0.7866

b) The probability of having at least 1 failure in a 24-hour day is 0.5132

Step-by-step explanation:

a) Lets call X the number of failures in a 8 hour period. Since it is 8 hours insstead of 1, the mean will be 8*0.030 = 0.24. As a consequence, the probability that there are no failures in this period (in other words, X = 0), is

b) Lets call Y the number of failures in 24 hours. Y has mean 24*0.03 = 0.72. The probability that Y is at least 1 can be computed by calculating the probability that Y is equal to 0 and substract from 1 this number (because obtaining 0 failures is the complementary event). The probability of Y being equal to 0 is

Thus, the probability of having at least 1 failure in a 24-hour day is 1-0.4868 = 0.5132

(a) 0.7866

(b) 0.5132

Step-by-step explanation:

For a Poisson random variable, X, its probability is given by

λ is the mean of the distribution and is given by

n is the number of hours in this question and p is the failure rate.

From the question, p = 0.030.

(a) The probability that the instrument does not fail in an 8-hour shift is the probability of no failure.

Here, n = 8. Then λ = 0.030 × 8 = 0.24

(b) The probability of at least 1 failure in a 24-hour day is the complement of no failure in 24 hours.

Here, n = 24. Then λ = 0.030 × 24 = 0.72