A teacher on her way to work passes through three stoplights

A teacher on her way to work passes through three stoplights each morning. The distances between the stop lights are great and the lights operate independently of each other. If the probablilities of a red light are .4,.8,and .6 respectively, for each light, what is the probability to the nearest hundredth that she will not have any red lights on her way to work?

Solution

probablilitiy of first red light = 0.4

probablilitiy of first light being not red = 1 - 0.4 = 0.6

probablilitiy of first red light = 0.8

probablilitiy of first light being not red = 1 - 0.8 = 0.2

probablilitiy of first red light = 0.6

probablilitiy of first light being not red = 1 - 0.6 = 0.4

probablilitiy that she will not have any red lights on her way to work = 0.6 * 0.2 * 0.4 = 0.048