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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 \(B, .98 ;\) and by both devices, .94 a. If smoke is present, find the probability that the smoke will be detected by device \(A\) or device \(B\) or both devices. b. Find the probability that the smoke will not be detected.

Short Answer

Expert verified
Answer: The probability of smoke being detected by either device A, device B, or both devices is 0.99, and the probability of smoke not being detected is 0.01.

Step by step solution

01

a. Finding the probability that smoke is detected by device A or device B or both devices.

Let's find the probability of detecting the smoke by device A or device B or both devices when smoke is present, using the formula for the probability of the union of two events: P(A U B) = P(A) + P(B) - P(A ∩ B) In our case, P(A) is the probability of detecting smoke by device A, P(B) is the probability of detecting smoke by device B, and P(A ∩ B) is the probability of detecting smoke by both devices. Plugging in the given values, we have: P(A U B) = 0.95 + 0.98 - 0.94 P(A U B) = 0.99 So, when smoke is present, the probability that smoke is detected by device A or device B, or both is \(0.99\).
02

b. Finding the probability that smoke is not detected

Now we want to find the probability that the smoke is not detected (i.e., the complement of the probability in part (a)). The probability of the complement event is given by: P(not detected) = 1 - P(detected) From part (a), we found the probability of smoke being detected by either device A, device B or both as 0.99. Thus, the probability of smoke not being detected is: P(not detected) = 1 - P(A U B) P(not detected) = 1 - 0.99 P(not detected) = 0.01 The probability that the smoke will not be detected is \(0.01\).

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Key Concepts

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

Union of Events in Probability
Understanding the union of events is essential for assessing the probability in scenarios involving multiple outcomes. In probability theory, the union of two events, say event A and event B, represents the set of outcomes that occur if either event A occurs, or event B occurs, or if they both occur simultaneously. One crucial aspect to grasp is that while summing the probabilities of two events, the probability of the intersection (the event they both occur) is often subtracted to avoid double-counting.

In mathematical terms, the probability of the union of two events, denoted as P(A U B), can be expressed using the formula:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] For the given exercise, wherein we calculate the probability of a smoke-detector system detecting smoke, we implement this formula. By finding out the individual probabilities of each device detecting smoke (device A and device B), and the probability of both detecting smoke together, we were able to calculate the overall likelihood of detection. This understanding is pivotal for problems where multiple interconnected events affect an outcome and is frequently applied across various disciplines, including statistical analysis and risk management.
Complement of an Event
Within the realms of probability theory, every event E has a complementary event, often denoted as E', which represents all outcomes not in E. Calculating the complement is a way to find the probability of 'what does not happen' if we know 'what does happen'. That is to say, if the probability of an event happening is known, the probability of the event not happening is simply one (the certainty of any outcome occurring) minus the probability of the event. Mathematically, this is illustrated as:
\[ P(E') = 1 - P(E) \] In our smoke-detector example, we first calculated the probability of smoke being detected. The complement of this event, or the probability of smoke not being detected, is calculated by subtracting this value from one. This method applies generally to situations where there are two mutually exclusive outcomes, such as pass/fail, on/off, or detected/not detected.
Probability Calculations
Probability calculations are integral to quantifying uncertainty and making informed decisions based on likely outcomes. These calculations involve applying formulas and principles of probability to data or given scenarios. As demonstrated in our exercise, we were able to deduce the likelihood of a smoke alarm detecting smoke through careful application of known probabilities.

For successful probability calculations, it's important to understand and correctly apply laws such as the addition rule for the union of events, the complement rule, and the multiplication rule for independent events, among others. Accuracy in computing these probabilities ensures reliable predictions and analyses in a wide array of fields, including science, engineering, finance, and daily decision-making. Always check assumptions, ensure values are within the range of 0 to 1 (as probabilities cannot exceed certainty), and remember that a probability of 0 indicates impossibility, whereas a probability of 1 indicates certainty.

Through exercises such as the one we explored, students build a solid foundation in assessing probabilities, which is invaluable for critical thinking and problem-solving in real-world applications.

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Most popular questions from this chapter

An investor has the option of investing in three of five recommended stocks. Unknown to her, only two will show a substantial profit within the next 5 years. If she selects the three stocks at random (giving every combination of three stocks an equal chance of selection), what is the probability that she selects the two profitable stocks? What is the probability that she selects only one of the two profitable stocks?

Many companies are testing prospective employees for drug use, with the intent of improving efficiency and reducing absenteeism, accidents, and theft. Opponents claim that this procedure is creating a class of unhirables and that some persons may be placed in this class because the tests themselves are not \(100 \%\) reliable. Suppose a company uses a test that is \(98 \%\) accurate \(-\) that is, it correctly identifies a person as a drug user or nonuser with probability \(98-\) and to reduce the chance of error, each job applicant is required to take two tests. If the outcomes of the two tests on the same person are independent events, what are the probabilities of these events? a. A nonuser fails both tests. b. A drug user is detected (i.e., he or she fails at least one test). c. A drug user passes both tests.

A random variable \(x\) has this probability distribution: $$\begin{array}{l|rrrrrr}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .1 & .3 & .4 & .1 & ? & .05\end{array}$$ a. Find \(p(4)\). b. Construct a probability histogram to describe \(p(x)\). c. Find \(\mu, \sigma^{2},\) and \(\sigma\). d. Locate the interval \(\mu \pm 2 \sigma\) on the \(x\) -axis of the histogram. What is the probability that \(x\) will fall into this interval? e. If you were to select a very large number of values of \(x\) from the population, would most fall into the interval \(\mu \pm 2 \sigma ?\) Explain.

An experiment is run as follows- the colors red, yellow, and blue are each flashed on a screen for a short period of time. A subject views the colors and is asked to choose the one he feels was flashed for the longest time. The experiment is repeated three times with the same subject. a. If all the colors were flashed for the same length of time, find the probability distribution for \(x\), the number of times that the subject chose the color red. Assume that his three choices are independent. b. Construct the probability histogram for the random variable \(x\)

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