Question

A contractor is required by a county planning department to submit one, two, three, four, five, or six forms (depending on the nature of the project) in applying for a building permit. Let Y = 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) = ky for y = 1, , 6. (Enter your answers as fractions.) (a) What is the value of k? [Hint: 6 y = 1 p(y) = 1] k = 1/21 (b) What is the probability that at most four forms are required? 10/21 (c) What is the probability that between two and five forms (inclusive) are required?

Answer 1

Answer:

a)

$k = \dfrac{1}{21}$

b) 0.476

c) 0.667    

Step-by-step explanation:

We are given the following in the question:

Y = the number of forms required of the next applicant.

Y: 1, 2, 3, 4, 5, 6

The probability is given by:

$P(y) = ky$

a) Property of discrete probability distribution:

$\displaystyle\sum P(y_i) = 1\\\\\Rightarrow k(1+2+3+4+5+6) = 1\\\\\Rightarrow k(21) = 1\\\\\Rightarrow k = \dfrac{1}{21}$

b) at most four forms are required

$P(y \leq 4) = \displaystyle\sum^{y=4}_{y=1}P(y_i)\\\\P(y \leq 4) = \dfrac{1}{21}(1+2+3+4) = \dfrac{10}{21} = 0.476$

c) probability that between two and five forms (inclusive) are required

$P(2\leq y \leq 5) = \displaystyle\sum^{y=5}_{y=2}P(y_i)\\\\P(2\leq y \leq 5) = \dfrac{1}{21}(2+3+4+5) = \dfrac{14}{21} = 0.667$