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Chapter 3: Problem 49

A parallel system functions whenever at least one of its components works. Consider a parallel system of \(n\) components and suppose that each component independently works with probability \(\frac{1}{2}\). Find the conditional probability that component 1 works given that the system is functioning.

Short Answer

Expert verified
The conditional probability that component 1 works given that the system is functioning is given by: $$ P(\text{Component 1 works} \mid \text{System is functioning}) = \frac{\frac{1}{2}}{1-\left(\frac{1}{2}\right)^n} $$

Step by step solution

01

Understanding Conditional Probability

Conditional probability is the probability of an event occurring, given that another event has already occurred. In our case, we want to find the probability that component 1 works, given that the system is functioning. This can be denoted as follows: $$ P(\text{Component 1 works} \mid \text{System is functioning}) $$
02

Using the Formula for Conditional Probability

The formula for conditional probability is: $$ P(A \mid B)=\frac{P(A \cap B)}{P(B)} $$ In this exercise, we have: - Event A: Component 1 works - Event B: System is functioning So, we want to find: $$ P(\text{Component 1 works} \mid \text{System is functioning}) = \frac{P(\text{Component 1 works} \cap \text{System is functioning)}}{P(\text{System is functioning})} $$
03

Finding the Probability of the Intersection

The system is functioning if at least one of its components is working. So the intersection of "Component 1 works" and "System is functioning" is just the event that Component 1 works since if Component 1 is working, it is implied that the system is functioning. Therefore, we have: $$ P(\text{Component 1 works} \cap \text{System is functioning}) = P(\text{Component 1 works}) = \frac{1}{2} $$
04

Finding the Probability of the System Functioning

We know that the system is functioning if at least one of its components is working. To find the probability of the system functioning, we can find the complementary probability that none of the n components works and then subtract it from 1. The probability of each component not working is also \(\frac{1}{2}\). And since the components work independently, we have: $$ P(\text{System not functioning}) = \frac{1}{2} \times \frac{1}{2} \times ... \times \frac{1}{2} = \left(\frac{1}{2}\right)^n $$ Now, we can find the probability of the system functioning: $$ P(\text{System is functioning}) = 1 - P(\text{System not functioning}) = 1 - \left(\frac{1}{2}\right)^n $$
05

Calculating the Conditional Probability

Now we can substitute these probabilities back into the conditional probability formula: $$ P(\text{Component 1 works} \mid \text{System is functioning}) = \frac{P(\text{Component 1 works} \cap \text{System is functioning)}}{ P(\text{System is functioning})} = \frac{\frac{1}{2}}{1-\left(\frac{1}{2}\right)^n} $$ Our answer is: $$ P(\text{Component 1 works} \mid \text{System is functioning}) = \frac{\frac{1}{2}}{1-\left(\frac{1}{2}\right)^n} $$

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Key Concepts

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

Parallel System Probability
Understanding how probability works in parallel systems is crucial for many engineering and reliability scenarios. A parallel system is one that functions as long as at least one of its components is working. When dealing with such systems, it's important to grasp that the total system's functionality is not dependent on each individual component. Instead, the focus is on ensuring that the system has at least one operational component.

For instance, if there are two components in a parallel system and each has a 50% chance of working (independently of the other), then the probability of both failing is only 25% (the product of their individual probabilities of failing). Thus, the system has a 75% chance of working because it will function if either or both of the components are functioning. This example demonstrates the power of redundancy in increasing system reliability.

The general formula to calculate the probability of a parallel system of independent components functioning is
\[ P(\text{System is functioning}) = 1 - (\text{probability that all components fail) }\] In a system where each component has an equal chance of failing, the formula simplifies to: \[ P(\text{System is functioning}) = 1 - (\text{individual component failure chance)}^n \] as seen in the exercise. This formula considers the complementary probability of system failure to deduce the system's functionality.
Independent Events
In probability theory, events are considered independent if the occurrence of one does not affect the probability of the occurrence of the other. This concept is pivotal to understanding many probability problems and is heavily relied upon when calculating the likelihood of outcomes of a system with multiple components.

For example, in the case of a parallel system, knowing that components are independent simplifies the analysis. If component A fails, this has no impact on the likelihood of component B working, provided the events are independent. As such, the probability of independent events A and B both occurring is the product of their individual probabilities: \[ P(A \cap B) = P(A) \times P(B) \]

However, it's important to note that not all events are independent. When events do affect each other's outcomes, they are referred to as dependent events, and different rules apply for calculating probabilities.
Complementary Probability
Complementary probability involves looking at the opposite side of the event of interest. In other words, it deals with the probability that the event does not occur. This concept is extremely useful in probability and statistics as it can often simplify complex problems.

For instance, when calculating the likelihood that a system functions, it can be easier to first find the chance that the system fails completely and then subtract this from 1. This reverse approach is precisely what the complementary probability is about. Mathematically, for any event A, the complementary probability is given by: \[ P(\text{not A}) = 1 - P(A) \] This method was illustrated in the exercise, where the probability that the system does not function (all components fail) had to be subtracted from 1 to get the chance that the system is functioning. Using the complementary probability is a classic technique to make the calculation more straightforward in scenarios where direct computation of an event's probability is complex.

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Most popular questions from this chapter

Let \(Q_{n}\) denote the probability that in \(n\) tosses of a fair coin no run of 3 consecutive heads appears. Show that $$ \begin{aligned} &Q_{n}=\frac{1}{2} Q_{n-1}+\frac{1}{4} Q_{n-2}+\frac{1}{8} Q_{n-3} \\ &Q_{0}=Q_{1}=Q_{2}=1 \end{aligned} $$ Find \(Q_{8}\). HINT: Condition on the first tail.

Three prisoners are informed by their jailer that one of them has been chosen at random to be executed, and the other two are to be freed. Prisoner \(A\) asks the jailer to tell him privately which of his fellow prisoners will be set free, claiming that there would be no harm in divulging this information because he already knows that at least one of the two will go free. The jailer refuses to answer this question, pointing out that if \(A\) knew which of his fellow prisoners were to be set free, then his own probability of being executed would rise from \(\frac{1}{3}\) to \(\frac{1}{2}\) because he would then be one of two prisoners. What do you think of the jailer's reasoning?

Barbara and Dianne go target shooting. Suppose that each of Barbara's shots hits the wooden duck target with probability \(p_{1}\), while each shot of Dianne's hits it with probability \(p_{2}\). Suppose that they shoot simultaneously at the same target. If the wooden duck is knocked over (indicating that it was hit), what is the probability that (a) both shots hit the duck; (b) Barbara's shot hit the duck? What independence assumptions have you made?

A certain organism possesses a pair of each of 5 different genes (which we will designate by the first 5 letters of the English alphabet). Each gene appears in 2 forms (which we designate by lowercase and capital letters). The capital letter will be assumed to be the dominant gene in the sense that if an organism possesses the gene pair \(x X\), then it will outwardly have the appearance of the \(X\) gene. For instance, if \(X\) stands for brown eyes and \(x\) for blue eyes, then an individual having either gene pair \(X X\) or \(x X\) will have brown eyes, whereas one having gene pair \(x x\) will have blue eyes. The characteristic appearance of an organism is called its phenotype, whereas its genetic constitution is called its genotype. (Thus 2 organisms with respective genotypes \(a A, b B, c c\), \(d D, e e\) and \(A A, B B, c c, D D, e e\) would have different genotypes but the same phenotype.) In a mating between 2 organisms each one contributes, at random, one of its gene pairs of each type. The 5 contributions of an organism (one of each of the 5 types) are assumed to be independent and are also independent of the contributions of its mate. In a mating between organisms having genotypes \(a A, b B, c C, d D, e E\) and \(a a, b B, c c, D d, e e\) what is the probability that the progeny will (i) phenotypically and (ii) genotypically resemble (a) the first parent; (b) the second parent; (c) either parent; (d) neither parent?

Consider 3 urns. Urn \(A\) contains 2 white and 4 red balls; urn \(B\) contains 8 white and 4 red balls; and um \(C\) contains 1 white and 3 red balls. If 1 ball is selected from each urn, what is the probability that the ball chosen from urn \(A\) was white, given that exactly 2 white balls were selected?
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