Mathematics: The number of failures of a testing instrument from contamination particles on the product is a Poisson random variable with a mean of 0.019 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 one failure in a 24-hour day?

The number of failures of a testing instrument from

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20.01.2022

a) 85.90% probability that the instrument does not fail in an 8-hour shift.

b) 36.62% probability of at least one failure in a 24-hour day

Step-by-step explanation:

The only information that we have is a mean during an interval. So we use the Poisson distribution solve this question.

We have that the probability of exactly x events is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*(\mu)^{x}}{x!}$

In which $\mu$ is the mean.

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

Mean of 0.019 failures per hour. For 8 hours

$\mu = 8*0.019 = 0.152$ .

This probability is P(X = 0). So

$P(X = x) = \frac{e^{-\mu}*(\mu)^{x}}{x!}$

$P(X = 0) = \frac{e^{-0.152}*(0.152)^{0}}{0!} = 0.8590$

85.90% probability that the instrument does not fail in an 8-hour shift.

(b) Probability of at least one failure in a 24-hour day?

Mean of 0.019 failures per hour. For 24 hours

$\mu = 24*0.019 = 0.456$ .

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

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

We want to find $P(X \geq 1)$ . So

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

In which

$P(X = x) = \frac{e^{-\mu}*(\mu)^{x}}{x!}$

$P(X = 0) = \frac{e^{-0.456}*(0.456)^{0}}{0!} = 0.6338$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6338 = 0.3662$

36.62% probability of at least one failure in a 24-hour day

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