/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 Traffic Light Problem 1: Two tra... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The menu at Paesano's lists 7 salads, 11 entrees, and 9 desserts. How many different salad-entree-dessert meals could you select? (Meals are considered to be different if any one thing is different.)

A salesperson has 7 customers in Denver and 13 customers in Reno. In how many different ways could she telephone a. A customer in Denver and then a customer in Reno? b. A customer in Denver or a customer in Reno, but not both?

Accident/Illness Insurance Problem: Some of the highest-paid mathematicians are the actuaries, who figure out what you should pay for various types of insurance. Suppose an insurance company has an accident/illness policy that pays 500 dollars if you get ill during any one year, 1000 dollars if you have an accident, and 6000 dollars if you both get ill and have an accident. The premium, or payment, for this policy is 100 dollars per year. One of your friends, who has studied actuarial science, tells you that your probability of becoming ill in any one year is 0.05 and that your probability of having an accident is 0.03. a. Find the probabilities of each event. i. Becoming ill and having an accident ii. Becoming ill and not having an accident iii. Not becoming ill but having an accident iv. Not becoming ill and not having an accident b. What is the mathematical expectation for this policy? c. An insurance policy is actuarially sound if the insurance company is expected to make a profit from it. Based on the probabilities assumed, is this policy actuarially sound?

Evaluate the number of combinations or permutations two ways: a. Using factorials, as in the examples of this section b. Directly, using the built-in features of your grapher $$6 C 4$$

Telephone Number Problem: When 10 -digit telephone numbers were introduced into the United States and Canada in the 1960 s, certain restrictions were placed on the groups of numbers: Area Code: 3 digits; the first must not be 0 or \(1,\) and the second must be 0 or 1 Exchange Code: 3 digits; the first and second must not be 0 or 1 Line Number: 4 digits; at least one must not be 0 a. Find the possible numbers of area codes, exchange codes, and line numbers. b. How many valid numbers could there be under this numbering scheme? c. How many 10-digit numbers could be made if there were no restrictions on the three groups of numbers? d. What is the probability that a 10 -digit number dialed at random would be a valid number under the original restrictions? e. The total population of the United States and Canada is currently about 300 million. In view of the fact that there are now area codes and exchange codes that do not conform to the original restrictions, what assumption can you make about the number of telephones per person in the United States and Canada?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.