Question

Data from the past shows that on average, a ready-mixed concrete plant receives 100 orders for concrete every year. The maximum number of orders that the plant can fulfill each week is 2. (a) What is the probability that in a given week the plant cannot fulfill all the placed orders? (b)Assume the answer to part (a) is 20% (It is not; I just want to make sure that everybody uses the same number for part (b)). Suppose there are 5 of such plants. What is the probability that in a given week 2 of the plants cannot fulfill their orders?

Answer 1

(a)

Let X be the number of orders received per week.

We can assume X to follow Poisson distribution as it possess below properties -

• The outcomes can be classified as successes (receiving an order) or failures (not receiving an order).
• The average number of successes that occurs in a specified time is known.
• The probability that a success will occur is proportional to the size of the time duration.
• The probability that a success will occur in an extremely small time is virtually zero.

Average number of orders per week = 100 * (7 / 365) = 1.918

X ~ Poisson(1.918)

Probability that in a given week the plant cannot fulfill all the placed orders = P(X > 2)

= 1 -( P(X = 0) + P(X = 1) + P(X = 2))

= 1 - (0.1469005 + 0.2817551 + 0.2702031)

= 0.3011

(b)

Let Y be number of plants that cannot fulfill their orders.

Y ~ Binomial(n = 5, p = 0.2)

Probability that in a given week 2 of the plants cannot fulfill their orders = P(Y = 2)

= 0.2048