Failures of a testing instrument from contamination particles on the product follow a Poisson process with a mean of 0.38 failures per an 8-hour shift. a) Find the probability that the instrument does not fail during a shift. b) Find the probability that there are more than 3 failures in a 48-hour period. c) Provided there is at least one failure in a 24-hour period, what is the probability that there are no more than 3 failures during that period.

(a) 0.62

(b) 0.2244

(c) 0.5097

Step-by-step explanation:

(a) The probability of a failure in a shift is : 0.38

The probability of no failure during a shift will thus be: 1 - 0.38 = 0.62

(b) Number of shifts in a 48 hour period = 48 / 8 = 6 shifts

Number of ways 4 failures could occur in 6 shifts (permutations with similar objects)  = 6! / (4!) = 24

Probability of 4 shifts with failures and 2 shifts without:

0.38 * 0.38 * 0.38 * 0.38 * 0.62 * 0.62 = 0.008

This means there is a total probability of 0.008 * 24 = 0.192 probability of 4 failures occurring in 6 shifts.

Similarly for 5 and 6 failures we calculate as follows:

Number of ways 5 failures can occur = 6! / 5! = 6

Probability of 5 failures = 0.38^5  *  0.62  = 0.0049

Total probability of 5 failures = 6 * 0.0049 = 0.0294

Number of ways 6 failures can occur = 1

Probability of 6 failures = 0.38^6 = 0.003

Adding probabilities of 4, 5 and 6 failures together we get the total probability for (b) = 0.192 + 0.0294 + 0.003 = 0.2244

(c) Since there is at least one failure in a 24 hour period, the number of failures in a 48 hour period must be 2.

Given this fact, let's calculate the probability of 3 failures. Here we will only consider the 4 shifts in which failure is not given already.

Probability of exactly 1 failure in the 4 shifts = 0.38 * 0.62^3 = 0.0905

We must also multiply this probability with the number of ways this can occur.

Number of ways this can occur : 4! / 3! = 4

So probability of 3 failures = 0.0905 * 4 = 0.362

The probability of only two failures = 0.62 ^ 4 = 0.1477

Total probability of no more than 3 failures given there are already two failures = 0.362 + 0.1477 = 0.5097