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Compute probability of no service days in seven day week

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Reference no: EM131674092

Question: For the system in Exercise I and Exercise II assume the plant works seven days a week. What is the probability the third service day occurs by the end of 10 days? Solve using the negative binomial distribution; repeat using the binomial distribution.

Exercise II: For the system in Exercise I, call a day in which one or more failures occur among the 350 lamps a "service day." Since a Bernoulli sequence "starts over" at any time, the sequence of service/ nonservice days may be considered a Bernoulli sequence with probability p1, the probability of one or more lamp failures in a day.

a. Beginning on a Monday morning, what is the probability the first service day is the first, second, third, fourth, fifth day of the week?

b. What is the probability of no service days in a seven day week?

Exercise I: A manufacturing plant has 350 special lamps on its production lines. On any day, each lamp could fail with probability p = 0:0017. These lamps are critical, and must be replaced as quickly as possible. It takes about one hour to replace a lamp, once it has failed. What is the probability that on any day the loss of production time due to lamp failaures is k or fewer hours, k = 0; 1; 2; 3; 4; 5?

Reference no: EM131674092

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