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Chapter 2: Problem 73

A boiler has five identical relief valves. The probability that any particular valve will open on demand is 95 . Assuming independent operation of the valves, calculate \(P\) (at least one valve opens) and \(P\) (at least one valve fails to open).

Short Answer

Expert verified
\(P(\text{at least one opens}) = 1 - (0.05)^5\) and \(P(\text{at least one fails}) = 1 - (0.95)^5\).

Step by step solution

01

Define Events

Let the event that any particular valve opens be denoted as an event with probability 0.95. Let the event where a particular valve does not open be denoted as the complement of the opening event, with probability 0.05 since this is 1 - 0.95.
02

Calculate Probability None Open

Calculate the probability that none of the valves open. Since the valves operate independently, the probability that none open is the product of the probabilities for each valve not opening. Thus, it is given by \((0.05)^5\).
03

Calculate Probability At Least One Opens

The probability that at least one valve opens is the complement of the probability that none of the valves open. Hence, it is calculated as \(1 - (0.05)^5\).
04

Calculate Probability At Least One Fails

Using similar logic, calculate the probability that at least one valve fails to open. This is the complement of the probability that all valves open. Thus, \((0.95)^5\) is the probability all valves open, and \(1 - (0.95)^5\) is the probability at least one fails to open.

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Key Concepts

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

Independent Events
In probability theory, understanding independent events is crucial for calculating complex probabilities. Independent events are those whose outcomes do not affect each other. For example, the functioning of each relief valve in the exercise is considered independent. This means the probability of one valve opening does not change the probability of another valve opening.
Such independence allows us to use multiplication to determine the overall probability of combined events.When you have several independent events, like our five valves, to find the probability that all events happen together, you multiply their individual probabilities. For instance, if each valve has a 0.95 probability to open, to calculate all valves opening you take \[ (0.95)^5 \]Understanding this concept simplifies multi-step probability calculations and helps in real-world scenarios where multiple independent actions are involved.
Complementary Probability
Complementary probability helps calculate the probability of an event not occurring by subtracting the probability of the event occurring from 1.Let's say you want to find the probability of a valve not opening in the exercise. If the probability of a valve opening is 0.95, then the probability of it not opening is simply: \[ 1 - 0.95 = 0.05 \]This principle becomes especially useful in cases where calculating the probability of none or some of the events happening is easier with complements. Just like in our exercise, to determine the probability that at least one valve opens, you can use: \[ 1 - (\text{probability no valves open}) = 1 - (0.05)^5 \]Using complementary probability can make some calculations easier and more intuitive, especially when dealing with events that encompass all possible outcomes.
Event Probability Calculation
Event probability calculation involves determining the chance of a specific outcome or set of outcomes occurring in a given scenario.
In this exercise, we look at two main events: at least one valve opens, and at least one valve fails.To start, calculating the probability that none of the valves open helps us find the complement; this involves using the probability of a single event (a valve not opening) multiplied across all independent valves:\[ (0.05)^5 \]To find out if at least one valve opens, use the complement:\[ 1 - (0.05)^5 \]Similarly, to calculate the probability that at least one valve fails, find the probability all valves open and subtract it from one:\[ 1 - (0.95)^5 \]Calculating event probabilities by breaking down into complements and combinations of independent events simplifies complex probabilities and provides insights into potential outcomes.

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Most popular questions from this chapter

A professional organization (for statisticians, of course) sells term life insurance and major medical insurance. Of those who have just life insurance, \(70 \%\) will renew next year, and \(80 \%\) of those with only a major medical policy will renew next year. However, \(90 \%\) of policyholders who have both types of policy will renew at least one of them next year. Of the policy holders \(75 \%\) have term life insurance, \(45 \%\) have major medical, and \(20 \%\) have both. a. Calculate the percentage of policyholders that will renew at least one policy next year. b. If a randomly selected policy holder does in fact renew next year, what is the probability that he or she has both life and major medical insurance?

A small manufacturing company will start operating a night shift. There are 20 machinists employed by the company. a. If a night crew consists of 3 machinists, how many different crews are possible? b. If the machinists are ranked \(1,2, \ldots, 20\) in order of competence, how many of these crews would not have the best machinist? c. How many of the crews would have at least 1 of the 10 best machinists? d. If one of these crews is selected at random to work on a particular night, what is the probability that the best machinist will not work that night?

A company uses three different assembly lines- \(A_{1}\), \(A_{2}\), and \(A_{3}\) - to manufacture a particular component. Of those manufactured by line \(A_{1}, 5 \%\) need rework to remedy a defect, whereas \(8 \%\) of \(A_{2}\) 's components need rework and \(10 \%\) of \(A_{3}\) 's need rework. Suppose that \(50 \%\) of all components are produced by line \(A_{1}\), \(30 \%\) are produced by line \(A_{2}\), and \(20 \%\) come from line \(A_{3}\). If a randomly selected component needs rework, what is the probability that it came from line \(A_{1}\) ? From line \(A_{2}\) ? From line \(A_{3}\) ?

Show that for any three events \(A, B\), and \(C\) with \(P(C)>0, P(A \cup B \mid C)=P(A \mid C)+P(B \mid C)-\) \(P(A \cap B \mid C)\).

A system consists of two components. The probability that the second component functions in a satisfactory manner during its design life is \(.9\), the probability that at least one of the two components does so is \(.96\), and the probability that both components do so is \(.75\). Given that the first component functions in a satisfactory manner throughout its design life, what is the probability that the second one does also?
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