Question

A contractor is required by a county planning department to submit anywhere from one to five forms (depending on the nature of the project) in applying for a building permit. Let \(y\) be the number of forms required of the next applicant. The probability that \(y\) forms are required is known to be proportional to \(y ;\) that is, \(p(y)=k y\) for \(y=1, \ldots, 5\) a. What is the value of \(k ?\) (Hint: \(\sum p(y)=1\) ) b. What is the probability that at most three forms are required? c. What is the probability that between two and four forms (inclusive) are required?

Short answer

a. The value of \(k\) is 1/15. b. The probability that at most three forms are required is 0.4. c. The probability that between two and four forms are required is 0.6.

Step-by-step solution

Find the value of \(k\)

To find the value of \(k\), we need to remember that the total probability for all possible outcomes should add up to 1. So we get \(\sum_{y=1}^{5} k \times y = 1\). This becomes \(k \times (1 + 2 + 3 + 4 + 5) = 1\). This is a simple arithmetic sum which gives \(k \times 15 = 1\). So, \(k = 1/15\).

Find the probability that at most three forms are required

This requires summing up the probabilities for \(y = 1\), \(y = 2\) and \(y = 3\). So \(P(Y \leq 3) = k \times (1 + 2 + 3) = 1/15 \times 6 = 0.4\).

Find the probability that between two and four forms are required

This involves summing up the probabilities for \(y = 2\), \(y = 3\) and \(y = 4\). So \(P(2 \leq Y \leq 4) = k \times (2 + 3 + 4) = 1/15 \times 9 = 0.6\).