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1 and #2. If one pump fails, the system will still operate. However, because of the added strain, the remaining pump is now m… # A system consists of two identical pumps, #1 and #2. If one pump fails, the system will still operate. However, because of the added strain, the remaining pump is now more likely to fail than was originally the case. That is, \(r=P(\\# 2\) fails \(\mid \\# 1\) fails \()>P(\\# 2\) fails \()=q\). If at least one pump fails by the end of the pump design life in \(7 \%\) of all systems and both pumps fail during that period in only \(1 \%\), what is the probability that pump #1 will fail during the pump design life?

Short Answer

Expert verified
The probability that pump #1 will fail is 0.04.

Step by step solution

01

Define the Known Probabilities

We are given two probabilities: \( P(A \cup B) = 0.07 \) and \( P(A \cap B) = 0.01 \), where \( A \) is the event that Pump #1 fails and \( B \) is the event that Pump #2 fails. We know that the probability of at least one pump failing is 7% and both pumps failing is 1%.
02

Use the Addition Rule for Probabilities

According to the addition rule for probabilities, \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \). We can rearrange this to find \( P(A) + P(B) = P(A \cup B) + P(A \cap B) = 0.07 + 0.01 = 0.08 \).
03

Recognize the Conditional Probability Inequality

We are told that \( r = P(B \mid A) > P(B) = q \). This means that if Pump #1 (event \( A \)) fails, Pump #2 (event \( B \)) is more likely to fail, compared to its standalone probability of failure.
04

Express \( P(B \mid A) \) Using its Definition

The conditional probability \( P(B \mid A) = \frac{P(A \cap B)}{P(A)} \). We know \( P(A \cap B) = 0.01 \), allowing us to set up the equation \( r = \frac{0.01}{P(A)} \).
05

Express \( P(B) \, \text{or} \, q \) in Terms of \( P(A) \)

From Step 2, we have \( P(A) + q = 0.08 \). Therefore, \( q = 0.08 - P(A) \). Substitute \( q \) and \( r \) into their inequality, given that \( r > q \).
06

Solve the Inequality

Substitute \( r = \frac{0.01}{P(A)} \) and \( q = 0.08 - P(A) \) into the inequality \( \frac{0.01}{P(A)} > 0.08 - P(A) \). Solve this inequality to find the interval for \( P(A) \).
07

Calculate \( P(A) \)

Solving the inequality \( \frac{0.01}{P(A)} > 0.08 - P(A) \) finds that \( P(A) = 0.04 \). Plugging this value back into our prior steps, we can verify that all conditions of the problem are satisfied.

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Key Concepts

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

Addition Rule for Probabilities
The addition rule for probabilities is a fundamental concept in probability theory. It helps us calculate the likelihood of either one of two events occurring. If you have two events, say Event A and Event B, the rule is expressed as follows:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]This equation states that the probability of either Event A or Event B occurring is equal to the sum of their individual probabilities, subtracting the probability that both events occur simultaneously. It accounts for the double-counting of the overlap where both events happen at the same time. Using this rule, we can find probabilities of combined events in situations like the pump system scenario where you want to know about any failures.
Conditional Probability
Conditional probability is used to determine the likelihood of an event occurring, given that another event has already occurred. In symbols, the conditional probability of Event B given Event A is represented as:\[ P(B \mid A) = \frac{P(A \cap B)}{P(A)} \]This formula shows us that the probability of Event B, given that Event A has already happened, is equal to the probability of both events happening together divided by the probability of Event A. It's a way to update probabilities based on new information. In our exercise, once Pump #1 fails, the probability that Pump #2 will also fail becomes higher. This is an application of conditional probability because the failure of Pump #2 is dependent on the failure of Pump #1.
System Reliability
System reliability refers to the ability of a system to function under specified conditions for a planned duration. In systems with components like pumps, reliability often means that the system continues to work even if one component fails, as long as others remain functional. Redundancies are designed to ensure continued operation. However, increased strain on remaining components, like a single operational pump having to do the work of two, may lead to higher chances of failure as seen in our problem. Understanding system reliability involves figuring out these interactions and making sure that the failure of one component does not lead to a total system breakdown. Such calculations ensure systems are designed to last, even when individual parts start failing.
Event Union and Intersection
Event union and intersection are basic operations in set theory applied to probabilities.- **Event Union (\(A \cup B\))**: Represents the occurrence of either Event A or Event B or both. Calculated using the addition rule, it signifies any conditions that lead to at least one of the events happening. - **Event Intersection (\(A \cap B\))**: Represents the occurrence of both Event A and Event B simultaneously. It's the shared space where both probabilities overlap.Understanding these concepts helps in complex probability scenarios, such as determining the occurrence of pump failures in our exercise. Calculating these intersections and unions is critical to solving problems where overlapping probabilities exist and can help predict the likelihood of multiple events occurring together.

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Most popular questions from this chapter

The three most popular options on a certain type of new car are a built-in GPS \((A)\), a sunroof \((B)\), and an automatic transmission \((C)\). If \(40 \%\) of all purchasers request \(A, 55 \%\) request \(B, 70 \%\) request \(C, 63 \%\) request \(A\) or \(B\), \(77 \%\) request \(A\) or \(C, 80 \%\) request \(B\) or \(C\), and \(85 \%\) request \(A\) or \(B\) or \(C\), determine the probabilities of the following events. [Hint: " \(A\) or \(B\) " is the event that at least one of the two options is requested; try drawing a Venn diagram and labeling all regions.] a. The next purchaser will request at least one of the three options. b. The next purchaser will select none of the three options. c. The next purchaser will request only an automatic transmission and not either of the other two options. d. The next purchaser will select exactly one of these three options.

Four universities-1,2, 3, and 4 -are participating in a holiday basketball tournament. In the first round, 1 will play 2 and 3 will play 4 . Then the two winners will play for the championship, and the two losers will also play. One possible outcome can be denoted by 1324 ( 1 beats 2 and 3 beats 4 in first- round games, and then 1 beats 3 and 2 beats 4 ). a. List all outcomes in \(\mathcal{S}\). b. Let \(A\) denote the event that 1 wins the tournament. List outcomes in \(A\). c. Let \(B\) denote the event that 2 gets into the championship game. List outcomes in \(B\). d. What are the outcomes in \(A \cup B\) and in \(A \cap B\) ? What are the outcomes in \(A^{\prime}\) ?

In five-card poker, a straight consists of five cards with adjacent denominations (e.g., 9 of clubs, 10 of hearts, jack of hearts, queen of spades, and king of clubs). Assuming that aces can be high or low, if you are dealt a five-card hand, what is the probability that it will be a straight with high card 10 ? What is the probability that it will be a straight? What is the probability that it will be a straight flush (all cards in the same suit)?

Show that if one event \(A\) is contained in another event \(B\) (i.e., \(A\) is a subset of \(B)\), then \(P(A) \leq P(B)\). [Hint: For such \(A\) and \(B, A\) and \(B \cap A^{\prime}\) are disjoint and \(B=\) \(A \cup\left(B \cap A^{\prime}\right)\), as can be seen from a Venn diagram.] For general \(A\) and \(B\), what does this imply about the relationship among \(P(A \cap B), P(A)\) and \(P(A \cup B)\) ?

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