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Traffic Light Problem 1: Two traffic lights on Broadway operate independently. Your probability of being stopped at the first light is \(40 \% .\) Your probability of being stopped at the second one is \(70 \% .\) Find the probability of being stopped at a. Both lights b. Neither light c. The first light but not the second d. The second light but not the first e. Exactly one of the lights

Short Answer

Expert verified
a) 0.28, b) 0.18, c) 0.12, d) 0.42, e) 0.54

Step by step solution

01

- Determine the Probability of Being Stopped at Both Lights

To find the probability of being stopped at both lights, multiply the probability of being stopped at the first light by the probability of being stopped at the second light. Probability of first light = 40%, or 0.4. Probability of second light = 70%, or 0.7. Thus, the combined probability is 0.4 * 0.7.
02

- Calculate the Probability of Not Being Stopped by Either Light

To calculate this probability, find the probability of NOT being stopped at each light and then multiply them together. The probability of not being stopped at the first light is 1 - 0.4 = 0.6, and at the second light is 1 - 0.7 = 0.3. Multiply these two probabilities together for the final answer.
03

- Find the Probability of Being Stopped at the First Light but Not the Second

First, take the probability of being stopped at the first light (0.4) and multiply it by the probability of not being stopped at the second light (0.3). This gives the probability for this scenario.
04

- Determine the Probability of Being Stopped at the Second Light but Not the First

To figure this out, multiply the probability of not being stopped at the first light (0.6) by the probability of being stopped at the second light (0.7).
05

- Compute the Probability of Being Stopped at Exactly One of the Lights

This probability is the sum of the probabilities found in Step 3 and Step 4. It will give the overall chance of being stopped at only one of the two traffic lights.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory
Probability theory is the branch of mathematics concerned with analyzing random events and quantifying the likelihood of various outcomes. It provides a framework for making predictions about systems and processes where chance plays a role. Understanding probability is essential for situating events in the context of uncertainty—like forecasting weather, assessing risk, or even determining the chances of getting stopped by traffic lights.

Consider the exercise where we calculate the probability of being stopped at traffic lights. Probability theory here allows students to ascertain the chances based on given percentages. Using decimal forms (0.4 and 0.7) simplifies multiplication, a common operation in probability calculations. It is crucial to comprehend that probabilities range from 0 (the event will not occur) to 1 (the event is certain to occur), with values in between representing the gradation of likelihood.
Independent Events
In probability, independent events are those whose outcomes do not affect each other. The probability of two independent events both occurring is the product of their individual probabilities.

In our traffic light problem, being stopped by the first light does not impact the probability of being stopped by the second light. They are independent. This is why, for part (a) of the exercise, the likelihood of being stopped at both lights can be found by simply multiplying the probabilities of each event occurring alone, i.e., 0.4 (first light) and 0.7 (second light). The outcome—being stopped at one light—does not modify the probability of being stopped at the other. This understanding is fundamental when solving problems involving multiple steps or stages that do not influence one another.
Complementary Probability
Complementary probability deals with the likelihood of an event not occurring, which is essential for understanding the full scope of possible outcomes. It can be found by subtracting the probability of the event from 1, because the sum of the probabilities of an event and its complement is always 1.

For instance, if the exercise asks for the probability of not being stopped at a light (part b), we compute the complement. If there is a 40% chance of being stopped at the first light, then there is a 60% chance (1 - 0.4) of not being stopped. Likewise, a 70% chance of being stopped at the second light means a 30% chance (1 - 0.7) of cruising through. To calculate the probability of passing both lights without stopping, we multiply their complementary probabilities together, showcasing their combined effect. This insight into complementary probability is critical for analyzing scenarios where the non-occurrence of an event is just as informative as its occurrence.

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Most popular questions from this chapter

Expectation of a Binomial Experiment: Suppose you conduct a random experiment that has a binomial probability distribution. Suppose the probability that outcome C occurs on any one repetition is \(0.4 .\) Let \(P(x)\) be the probability that outcome C occurs \(x\) times in five repetitions. a. Calculate \(P(x)\) for each value of \(x\) in the domain. b. Find the mathematically expected value of \(x\) (Hint: The value if \(C\) occurs \(x\) times is \(x .\) ) c. Show that the mathematically expected value of \(x\) is equal to 0.4 (the probability \(C\) occurs on one repetition) times 5 (the total number of repetitions). d. If the probability that \(C\) occurs on any one repetition is \(b\) and the probability that \(C\) does not occur on one repetition is \(a=1-b,\) prove that in five trials, the expected value of \(x\) is \(5 b\) e. From what you have observed in this problem, make a conjecture about the mathematically expected value of \(x\) in \(n\) repetitions, if the probability that C occurs on any one repetition is \(b\) f. If you plant 100 seeds, each of which has a probability of 0.71 of germinating, how many seeds would you expect to germinate?

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