/*! 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 94 A smoke detector system uses two... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A smoke detector system uses two devices, \(A\) and \(B\). If smoke is present, the probability that it will be detected by device \(A\) is \(.95 ;\) by device \(B, .90 ;\) and by both devices, .88 . a. If smoke is present, find the probability that the smoke will be detected by either device \(A\) or \(B\) or both devices. b. Find the probability that the smoke will be undetected.

Short Answer

Expert verified
a. 0.97; b. 0.03

Step by step solution

01

Understand the Problem

We need to find the probability that the smoke will be detected by either device A, device B, or both. Then, we need to calculate the probability that smoke will not be detected by any device.
02

Use the Probability Formula for Either Or Both

To find the probability that smoke is detected by either A or B or both, use the formula for the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]Where \(P(A) = 0.95\), \(P(B) = 0.90\), and \(P(A \cap B) = 0.88\).
03

Calculate Probability of Detection

Substitute the given probabilities into the formula: \[P(A \cup B) = 0.95 + 0.90 - 0.88 = 0.97\]Thus, the probability that smoke is detected by either device or both is 0.97.
04

Calculate the Probability of Non-Detection

To find the probability that smoke is not detected by either device, we use the complement rule: \[P( ext{undetected}) = 1 - P(A \cup B)\]Simplify to get:\[P( ext{undetected}) = 1 - 0.97 = 0.03\]
05

Concluding Step: Verify Result

Check our calculations: The sum of the probability of detection and non-detection should equal 1. Here, \(0.97 + 0.03 = 1\), which confirms our calculations are correct.

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.

Union of Events
In probability theory, the concept of the 'Union of Events' helps us determine the likelihood that at least one of several events occurs. It is particularly useful when considering multiple events that could happen simultaneously or independently, such as detecting smoke with two separate devices, as in our exercise.
The union of events is represented using the symbol \( \cup \), and to calculate the probability of either event A or B or both occurring, we use the formula:
  • \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
This formula takes into account the overlap or intersection \( P(A \cap B) \) that could happen when both events occur at the same time. By subtracting this overlap, which is counted twice in the addition \( P(A) + P(B) \), we get an accurate probability of the union of events.
In the case of smoke detection, it means calculating the chance that at least one of the devices detects smoke.
Complement Rule
The Complement Rule in probability is an essential tool for finding the likelihood of an event not occurring. When combined with the probability of an event occurring, these two probabilities always sum up to 1, representing certainty in probability terms.
The rule is expressed as:
  • \[ P(\text{not } A) = 1 - P(A) \]
This formula calculates the probability of the complement of event A (whatever is not A). In our smoke detection scenario, after calculating the union of events (detection by either or both devices), we use the complement rule to find the probability of non-detection. Since \( P(A \cup B) = 0.97 \), using the complement rule gives us \( P(\text{undetected}) = 1 - 0.97 = 0.03 \). This means there's a 3% chance that neither device detects smoke.
Event Probability
Event probability measures how likely a specific event is to occur within a given set of circumstances. In statistics, each individual event has its probability, a number between 0 (impossible event) and 1 (certain event).
In our example, device A has a probability of 0.95 to detect smoke, and device B has a probability of 0.90. These numbers indicate how effective each device is independently.
Understanding individual event probabilities assists in determining more complex probabilities, like the union of events. When various independent and dependent events contribute to the outcome, each event's probability is a crucial building block to compute overall likelihoods.
Detection and Non-Detection
In practical applications such as smoke detection, establishing the probability of detection and non-detection is vital for designing efficient systems.
Detection refers to the system's ability to identify a specific event, like smoke presence. In our exercise, we calculated that the smoke is detected with a high probability of 0.97 by employing two devices collaboratively.
Non-detection is equally important to consider, as it reflects the system's failure rate. By calculating non-detection probabilities, which was 0.03 in our example, people can assess risks associated with failing to detect smoke. Recognizing the balance between detection and non-detection can lead to improved systems and decision-making.

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

A study of Georgia residents suggests that those who worked in shipyards during World War were subjected to a significantly higher risk of lung cancer (Wall Street Journal, September 21, 1978). \(^{\star}\) It was found that approximately \(22 \%\) of those persons who had lung cancer worked at some prior time in a shipyard. In contrast, only \(14 \%\) of those who had no lung cancer worked at some prior time in a shipyard. Suppose that the proportion of all Georgians living during World War II who have or will have contracted lung cancer is \(.04 \% .\) Find the percentage of Georgians living during the same period who will contract (or have contracted) lung cancer, given that they have at some prior time worked in a shipyard.

Suppose that \(A\) and \(B\) are two events such that \(P(A)+P(B)<1\). a. What is the smallest possible value for \(P(A \cap B) ?\) b. What is the largest possible value for \(P(A \cap B) ?\)

A brand of automobile comes in five different styles, with four types of engines, with two types of transmissions, and in eight colors. a. How many autos would a dealer have to stock if he included one for each style-engine-transmission combination? b. How many would a distribution center have to carry if all colors of cars were stocked for each combination in part (a)?

According to Webster's New Collegiate Dictionary, a divining rod is "a forked rod believed to indicate [divine] the presence of water or minerals by dipping downward when held over a vein." To test the claims of a divining rod expert, skeptics bury four cans in the ground, two empty and two filled with water. The expert is led to the four cans and told that two contain water. He uses the divining rod to test each of the four cans and decide which two contain water. a. List the sample space for this experiment. b. If the divining rod is completely useless for locating water, what is the probability that the expert will correctly identify (by guessing) both of the cans containing water?

Americans can be quite suspicious, especially when it comes to government conspiracies. On the question of whether the U.S. Air Force has withheld proof of the existence of intelligent life on other planets, the proportions of Americans with varying opinions are given in the table. $$\begin{array}{lc} \hline \text { Opinion } & \text { Proportion } \\ \hline \text { Very likely } & .24 \\ \text { Somewhat likely } & .24 \\ \text { Unlikely } & .40 \\ \text { Other } & .12 \\ \hline \end{array}$$ Suppose that one American is selected and his or her opinion is recorded. a. What are the simple events for this experiment? b. Are the simple events that you gave in part (a) all equally likely? If not, what are the probabilities that should be assigned to each? c. What is the probability that the person selected finds it at least somewhat likely that the Air Force is withholding information about intelligent life on other planets?

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.