a teacher on her way to work passes through 3 stoplights each morning. the distances between the stop lights are great and the lights operate independently of each other. if the probability of the red lights are .4,.8 and .6 for each light, what is the probability to the nearest hundredth that she wont have any red lights on her way to work?

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

(1-.4)(1-.8)(1-.6) = ?

To find the probability that the teacher won't have any red lights on her way to work, we need to determine the probability of each stoplight being green and then multiply those probabilities together.

Let's calculate the probability for each stoplight:

1. Probability of the first stoplight being green: P(G1) = 1 - P(R1) = 1 - 0.4 = 0.6
2. Probability of the second stoplight being green: P(G2) = 1 - P(R2) = 1 - 0.8 = 0.2
3. Probability of the third stoplight being green: P(G3) = 1 - P(R3) = 1 - 0.6 = 0.4

Now, let's multiply these probabilities together to find the probability of all three stoplights being green:

P(No red lights) = P(G1) * P(G2) * P(G3) = 0.6 * 0.2 * 0.4 = 0.048

Therefore, the probability to the nearest hundredth that the teacher won't have any red lights on her way to work is 0.05.