Question

A factory produces 3 diesel-generator sets per week. At the end of each week, the sets are tested. If the sets are acceptable, they are shipped to purchasers. The probability that a set proves to be acceptable is 0.70. The second possibility is that minor adjustments can be made so that a set will become acceptable for shipping; this has a probability of 0.20. The third possible outcome is that the set has to go to the diagnostic shop for major adjustment and be shipped at a later date; this has a probability of 0.10. Outcomes for different sets are independent of one another. (a) Find the probability of each possible number of sets, for one week’s production, which are acceptable without any adjustment. (b) What is the expected number of sets which are tested and found to be acceptable without adjustment? (c) What is the cumulative probability distribution for the number of sets which are tested and found to be acceptable without adjustment? Sketch the corresponding graph​

Answer 1

(a) The probability of each possible number of sets that are acceptable without any adjustment can be found using the binomial distribution formula:

P(x) = (n choose x) * p^x * (1-p)^(n-x)

Where:

P(x) is the probability of x successes in n trials

n is the number of trials (in this case, the number of sets produced in a week)

x is the number of successes (in this case, the number of sets that are acceptable without any adjustment)

p is the probability of success (in this case, the probability that a set is acceptable without any adjustment, which is 0.70)

Plugging in the values, we get:

P(0) = (3 choose 0) * 0.70^0 * (1-0.70)^(3-0) = 0.028

P(1) = (3 choose 1) * 0.70^1 * (1-0.70)^(3-1) = 0.081

P(2) = (3 choose 2) * 0.70^2 * (1-0.70)^(3-2) = 0.162

P(3) = (3 choose 3) * 0.70^3 * (1-0.70)^(3-3) = 0.126

(b) To find the expected number of sets that are tested and found to be acceptable without adjustment, we can use the formula:

E(x) = n * p

Plugging in the values, we get:

E(x) = 3 * 0.70 = 2.1

The expected number of sets that are tested and found to be acceptable without adjustment is 2.1.

(c) The cumulative probability distribution for the number of sets that are tested and found to be acceptable without adjustment is the sum of the probabilities of each possible number of successes. For example, the probability of 0 or 1 successes is the sum of the probabilities of 0 and 1 successes: P(0 or 1) = P(0) + P(1) = 0.028 + 0.081 = 0.109. The probability of 0, 1, or 2 successes is the sum of the probabilities of 0, 1, and 2 successes: P(0 or 1 or 2) = P(0) + P(1) + P(2) = 0.028 + 0.081 + 0.162 = 0.271.

The cumulative probability distribution for the number of sets that are tested and found to be acceptable without adjustment is:

P(0) = 0.028

P(0 or 1) = P(0) + P(1) = 0.028 + 0.081 = 0.109

P(0 or 1 or 2) = P(0) + P(1) + P(2) = 0.028 + 0.081 + 0.162 = 0.271

P(0 or 1 or 2 or 3) = P(0) + P(1) + P(2) + P(3) = 0.028 + 0.081 + 0.162 + 0.126 = 0.397

The corresponding graph would be a step function with a step at each possible value of x (in this case, 0, 1, 2, and 3) and the corresponding probability at each step.