Comments on: Intro to Statistics Help!!!? http://sportcarx.com/2010/03/intro-to-statistics-help/ Sport Cars and Automotive Update Wed, 07 Jul 2010 21:20:59 -0700 http://wordpress.org/?v=2.8.6 hourly 1 By: amgasiago http://sportcarx.com/2010/03/intro-to-statistics-help/comment-page-1/#comment-701 amgasiago Wed, 17 Mar 2010 10:41:24 +0000 http://sportcarx.com/2010/03/intro-to-statistics-help/#comment-701 assuming each person's decision is independent of the decisions of all other people: to solve for the probably of non of the four renting a van, first find the probability of someone renting a van, which is .24, and the probability of them not renting a van is 1-.24, which is .76, so to get the probability of all four people not renting a van, put .76^4 = (.76)(.76)(.76)(.76) to solve for the probability that at least one of the four rents a van, take 1-P(none of the renting a van) = 1-(your previous answer) = 1-[(.76)(.76)(.76)(.76)] to solve for the probability that all three rent an SUV, take the probility that one rents an SUV, .12 and put that to the third power, or (.12)(.12)(.12) assuming each person’s decision is independent of the decisions of all other people:

to solve for the probably of non of the four renting a van, first find the probability of someone renting a van, which is .24, and the probability of them not renting a van is 1-.24, which is .76, so to get the probability of all four people not renting a van, put .76^4 = (.76)(.76)(.76)(.76)

to solve for the probability that at least one of the four rents a van, take 1-P(none of the renting a van) = 1-(your previous answer) = 1-[(.76)(.76)(.76)(.76)]

to solve for the probability that all three rent an SUV, take the probility that one rents an SUV, .12 and put that to the third power, or (.12)(.12)(.12)

]]>