/*! 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 25 There are two traffic lights on ... [FREE SOLUTION] | 91Ó°ÊÓ

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There are two traffic lights on Shelly's route from home to work. Let \(E\) denote the event that Shelly must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=0.4, P(F)=0.3\), and \(P(E \cap F)=0.15\). a. Use the given probability information to set up a hypothetical 1000 table with columns corresponding to \(E\) and not \(E\) and rows corresponding to \(F\) and not \(F\). b. Use the table from Part (a) to find the following probabilities: i. the probability that Shelly must stop for at least one light (the probability of \(E \cup F)\). ii. the probability that Shelly does not have to stop at either light. iii. the probability that Shelly must stop at exactly one of the two lights. iv. the probability that Shelly must stop only at the first light.

Short Answer

Expert verified
a. Hypothetical table: | | E | Not E | |---|-----|-------| | F | 150 | 150 | |Not F | 250 | 450 | b. Probabilities: i. \(P(E \cup F) = \frac{550}{1000} = 0.55\) ii. \(P(\sim E \cap \sim F) = \frac{450}{1000} = 0.45\) iii. \(P(\textrm{exactly one light}) = \frac{400}{1000} = 0.4\) iv. \(P(\textrm{only at the first light}) = \frac{250}{1000} = 0.25\)

Step by step solution

01

Calculate probabilities based on 1000 events

To set up the table, we will first determine the number of events by multiplying probabilities by 1000: - Stopping at both lights (E ∩ F): 0.15 * 1000 = 150 - Stopping at the first light only (E): (0.4 - 0.15) * 1000 = 250 - Stopping at the second light only (F): (0.3 - 0.15) * 1000 = 150 - Not stopping at any light (not E and not F): 1000 - 150 - 250 - 150 = 450 Now we can set up the table: | | E | Not E | |---|-----|-------| | F | 150 | 150 | |Not F | 250 | 450 | b. Find various probabilities: i. The probability that Shelly must stop for at least one light (E ∪ F)
02

Calculate the probability of E ∪ F

For this probability, we add the probabilities of the three non-empty cells according to the table: 150 + 150 + 250 = 550. Then we divide by 1000: 550/1000 = 0.55. ii. The probability that Shelly does not have to stop at either light
03

Calculate the probability of not E and not F

From the table, we can directly find this probability by dividing the value of the event where Shelly stops at neither light (450) by the total number of events (1000). So the probability is 450/1000 = 0.45. iii. The probability that Shelly must stop at exactly one of the two lights
04

Calculate the probability of stopping at exactly one light

To find this probability, we add the probabilities for stopping at E only and F only. From the table, this would be 150 (F and not E) + 250 (not F and E) = 400. Then we divide by 1000: 400/1000 = 0.4. iv. The probability that Shelly must stop only at the first light
05

Calculate the probability of stopping only at E

To find this probability, we divide the value of the event where Shelly stops at E and not F (250) by the total number of events (1000). So the probability is 250/1000 = 0.25.

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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 mathematical framework for quantifying the likelihood of various outcomes in an uncertain environment. It provides a set of principles and methods to calculate how probable an event is. In the context of our traffic light exercise, we use this theoretical basis to assess the chances of different combinations of stopping or not stopping at traffic lights.

For instance, Shelly's chance of having to stop at the first light () is 0.4, meaning there is a 40% likelihood of this event occurring. We are able to set up scenarios and calculate probabilities because probability theory assumes certain axioms. One such axiom is the , which states that the sum of the probabilities of all possible outcomes in a given scenario equals 1. This rule was used when calculating the probability of not having to stop at either light by subtracting the other probabilities from 1.
Conditional Probability
Conditional probability considers the likelihood of an event occurring, given that another event has already occurred. This concept greatly enhances our understanding of the dependence between events.

In our exercise, the probability of Shelly stopping at both lights () is a form of conditional probability. Here, we consider the second event occurring under the condition that the first one has already taken place. Calculating the conditional probability helps us understand dependencies among events, which is paramount in more complex situations than the two-traffic-light scenario for Shelly. For example, if we wanted to find the likelihood of Shelly stopping at the second light given she's already stopped at the first, we would look at the ratio of the intersection () over the probability of the first event ().
Probability Models
Probability models are mathematical representations of complex real-world processes that help us predict and analyze outcomes. A model might range from something as simple as a coin toss to the intricate systems of financial markets.

In the given exercise, we've constructed a basic probability model with a table that reflects Shelly's traffic light problem. Each cell within the table represents a distinct event with an associated probability, creating a visual and mathematical representation of all possible outcomes. Enhancing the table to include a hypothetical 1000 events eases the computation and is a part of the modeling process. Probability models are indispensable tools in scenarios ranging from daily occurrences, like traffic patterns, to life-saving applications, such as predicting the spread of diseases in epidemiology.

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Most popular questions from this chapter

An appliance manufacturer offers extended warranties on its washers and dryers. Based on past sales, the manufacturer reports that of customers buying both a washer and a dryer, \(52 \%\) purchase the extended warranty for the washer, \(47 \%\) purchase the extended warranty for the dryer, and \(59 \%\) purchase at least one of the two extended warranties. a. Use the given probability information to set up a hypothetical 1000 table. b. Use the table from Part (a) to find the following probabilities: i. the probability that a randomly selected customer who buys a washer and a dryer purchases an extended warranty for both the washer and the dryer. ii. the probability that a randomly selected customer purchases an extended warranty for neither the washer nor the dryer.

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