Question

A contractor is planning the acquisition of bulldozers needed for a new project in a remote area. From prior experience with this equipment he estimates that there is a 60% chance that each bulldozer will be operational for at least six months. If he purchases four bulldozers, it is desired to calculate the probabilities that exactly: 0, 1, 2, 3, and 4 bulldozers will be operational at the end of six months. Please document the following information:
a. Count the number of possible outcomes of the status of each bulldozer at the end of six months:
b. List the five events listed above as a function of the status of each bulldozer at the end of six months documented in 1.); and
c. Calculate the probabilities of these events assuming that the status of each bulldozer is statistically independent of the status of the rest of bulldozers.

Answer 1

Answer:

0.0256 ; 0.1536 ; 0.3456 ; 0.3456 ; 0.1296

Step-by-step explanation:

P, that each bulldozer will be operational for atleast 6 months = 0.6

p = 0.6 ; q = 1 - p = 1 - 0.6 = 0.4

Number of bulldozers purchased, n = 4

Probability that :

Exactly 0 bulldozer will be operational at the end of 6 months :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 0) = 4C0 * 0.6^0 * 0.4^4

P(x = 0) = 1 * 1 * 0.0256

P(x = 0) = 0.0256

Probability that :

Exactly 1 bulldozer will be operational at the end of 6 months :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 1) = 4C1 * 0.6^1 * 0.4^3

P(x = 1) = 4 * 0.6 * 0.064

P(x = 1) = 0.1536

Probability that :

Exactly 2 bulldozer will be operational at the end of 6 months :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 2) = 4C2 * 0.6^2 * 0.4^2

P(x = 2) = 6 * 0.36 * 0.16

P(x = 2) = 0.3456

Probability that :

Exactly 3 bulldozer will be operational at the end of 6 months :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 3) = 4C3 * 0.6^3 * 0.4^1

P(x = 3) = 4 * 0.216 * 0.4

P(x = 3) = 0.3456

Probability that :

Exactly 4 bulldozer will be operational at the end of 6 months :

P(x = x) = nCx * p^x * q^(n-x)

P(x = 4) = 4C4 * 0.6^4 * 0.4^0

P(x = 1) = 1 * 0.1296 * 1

P(x = 1) = 0.1296