Answer to Question #133584 in Statistics and Probability for miko

Question

A study was done by an undergraduate student wanting to determine the probable location of computers (PC or laptop) in the home. The following probabilities were determined: Adult bedroom = 0.03, Child bedroom = 0.10, Other bedroom = 0.14, Home Office or Study room = 0.42, Other rooms = 0.31. What is the probability that a computer is not in the study?

Answer 1

Given that the probability P(s) of location of the laptop in the Study room = 0.42

Hence the probability P that the laptop is not in the study room = 1 - probability of location in the study room.

P = 1 - P(s) = 1 - 0.42 = 0.58

P = 0.58