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A shipment of 5000 printed circuit boards contains 40 that are defective. Two boards will be chosen at random, without replacement. Consider the two events \(E_{1}=\) event that the first board selected is defective and \(E_{2}=\) event that the second board selected is defective. a. Are \(E_{1}\) and \(E_{2}\) dependent events? Explain in words. b. Let \(n o t E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(E_{1}\) )? c. How do the two probabilities \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid\right.\) not \(\left.E_{1}\right)\) compare? d. Based on your answer to Part (c), would it be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent?

Short Answer

Expert verified
a) Yes, E1 and E2 are dependent events. b) \(P(E1^C) = 0.992\). c) \(P(E2 | E1)\) and \(P(E2 | E1^C)\) are almost equal, but not quite. d) No, it is not reasonable to view E1 and E2 as approximately independent, despite the slight difference in probabilities.

Step by step solution

01

Determine dependence

Event E1 and E2 are dependent events because the selection of a defective board (E1) affects the probability of the next board being defective (E2). This is due to the fact that the boards are chosen without replacement, reducing the total number of boards available for the second draw.
02

Calculate the probability of the complement of E1

To calculate \(P(E1^C)\), the probability of the first board selected is not defective, we need to know the total number of boards and the number of defective boards. The total number of boards is 5000, and the number of defective boards is 40. So there are 5000 - 40 = 4960 non-defective boards. Thus \(P(E1^C) = \frac{4960}{5000} = 0.992\).
03

Compare conditional probabilities

The conditional probability \(P(E2 | E1)\), which is the probability of the second board being defective given that the first board is defective, is computed by \(P(E2 | E1) = \frac{39}{4999}\) (since after a defective PCB is picked, there are 39 defective PCB left out of 4999 total PCBs).The conditional probability \(P(E2 | E1^C)\), which is the probability of the second board being defective given that the first board is not defective, is computed by \(P(E2 | E1^C) = \frac{40}{4999}\) (since a non-defective PCB is picked first, there are still 40 defective PCBs left out of 4999 total PCBs). These two probabilities are almost but not quite equal.
04

Evaluate independence

Given that \(P(E2 | E1)\) and \(P(E2 | E1^C)\) are nearly equal, one might think that the events E1 and E2 are approximately independent. However, approximately independent is not a recognized condition in probability theory. E1 and E2 are either dependent or independent, and as identified in Step 1, they are dependent events.

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Key Concepts

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

Dependent Events
When discussing probability, dependent events are situations where the occurrence of one event affects the probability of another event happening. In the example of selecting printed circuit boards, events \(E_1\) and \(E_2\) are examples of dependent events. Since the boards are selected without replacement, choosing a defective board first reduces the total number of defective boards available for the second selection.
This dependency alters the likelihood of the second event since the pool of remaining boards changes after the first selection. It is crucial to consider this in your calculations, as standard probability formulas assume independent events unless otherwise specified.
  • In cases where items are not replaced, like our circuit boards, knowing the outcome of one selection changes the probability landscape.
  • In practical terms, dependent events are common in real-world scenarios, highlighting the importance of understanding how one outcome can influence another.
Always clearly distinguish between independent and dependent events to ensure correct probability interpretations.
Conditional Probability
Conditional probability is a powerful concept that allows us to calculate the likelihood of an event based on the occurrence of another related event. It is often denoted as \(P(A \mid B)\), which reads as "the probability of A given B". In our example, calculating \(P(E_2 \mid E_1)\) helps us understand the probability of the second circuit board being defective, provided the first board was defective.
We derived \(P(E_2 \mid E_1) = \frac{39}{4999}\) because one defective board was already chosen, leaving 39 defective ones among fewer total boards.
  • Conditional probability reshapes our probability space and accounts for new information or events that have already occurred.
  • It's highly useful for calculating probabilities in complex situations where outcomes are not isolated.
Additionally, understanding \(P(E_2\mid E_1^C)\), where the first board is not defective, illustrates how different scenarios should be separately accounted for. Here, \(P(E_2\mid E_1^C) = \frac{40}{4999}\) due to no initial decrease in defective boards.
Independence in Probability
Independence in probability indicates that two events have no influence over each other's outcomes. When two events are independent, the occurrence of one does not change the probability of the other occurring. Evaluating this within the context of selecting circuit boards provides valuable insights.
In our problem, even though \(P(E_2 \mid E_1)\) and \(P(E_2 \mid E_1^C)\) seem close in value, they aren't equal, leading to the conclusion that \(E_1\) and \(E_2\) are not independent.
  • True independence would mean that the conditional probabilities are the same irrespective of the occurrence of other events.
  • This highlights the importance of scrutinizing assumptions of independence as calculations heavily rely on these differentiations.
Recognizing independence or dependence is essential for accurately interpreting real-world situations and formulating correct probability models.

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Most popular questions from this chapter

Consider the following information about travelers on vacation: \(40 \%\) check work e-mail, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them on vacation, \(23 \%\) both check work e-mail and use a cell phone to stay connected, and \(51 \%\) neither check work e-mail nor use a cell phone to stay connected nor bring a laptop. In addition \(88 \%\) of those who bring a laptop also check work e-mail and \(70 \%\) of those who use a cell phone to stay connected also bring a laptop. With \(E=\) event that a traveler on vacation checks work e-mail, \(C=\) event that a traveler on vacation uses a cell phone to stay connected, and \(L=\) event that a traveler on vacation brought a laptop, use the given information to determine the following probabilities. A Venn diagram may help. a. \(P(E)\) b. \(P(C)\) c. \(P(L)\) d. \(P(E\) and \(C)\) e. \(P\left(E^{C}\right.\) and \(C^{C}\) and \(\left.L^{C}\right)\) f. \(P(E\) and \(\operatorname{Cor} L)\) g. \(\quad P(E \mid L)\) h. \(P(L \mid C)\) i. \(P(E\) and \(C\) and \(L)\) j. \(P(E\) and \(L)\) k. \(P(C\) and \(L)\) 1\. \(P(C \mid E\) and \(L)\)

Define the term chance experiment, and give an example of a chance experiment with four possible outcomes.

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