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Chapter 6: Problem 55

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 not \(E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(\left.E_{1}\right)\) ? 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, \(E_{1}\) and \(E_{2}\) are technically dependent events in this context, as selection of first board affects the composition of the boards that the second board is chosen from. But since number of defective is much less than total, the dependence is negligible.\n b. The probability of not \(E_{1}\) is 0.992.\n c. The two conditional probabilities, \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid E_{1}^{C}\right)\), are very similar.\n d. Given the similarity of the two conditional probabilities, \(E_{1}\) and \(E_{2}\) could be viewed as approximately independent.

Step by step solution

01

Dependence of events

Events \(E_{1}\) and \(E_{2}\) are dependent. This is because once a board is selected, the total number of boards reduces by 1. Hence, the probability of the second event depends on the outcome of the first event: if the first board is defective, there would be 39 defective boards left out of 4999, if not, there would still be 40 defective boards out of 4999.
02

Probability of not \(E_{1}\)

The event \(E_{1}^{C}\), or not \(E_{1}\), is that the first board selected is not defective. There are 5000-40=4960 such favourable cases. Hence, \(P\left(E_{1}^{C}\right) = \frac{4960}{5000} = 0.992\).
03

Comparison of the conditional probabilities

The conditional probability \(P\left(E_{2} \mid E_{1}\right) = \frac{39}{4999}\) because if the first selected board is defective (event \(E_{1}\)), we have 39 defective boards remaining out of 4999. The conditional probability \(P\left(E_{2} \mid E_{1}^{C}\right) = \frac{40}{4999}\) because if the first selected board is not defective, we still have 40 defective boards remaining out of 4999. These two probabilities are quite similar.
04

Independent or dependent

Even though the events are technically dependent, the two probabilities calculated in step 3 are very similar. This is due to the fact that the number of defective boards is very small compared to the total number of boards. Hence, it would be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent.

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Key Concepts

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

Dependent Events
In probability, events are considered dependent if the outcome of one event affects the probability of the other event. This is like our printed circuit board example, where selecting the first board changes the probabilities for the second. Once you select one board, the total number of boards available for the second selection is reduced by one. This shift in total affects the likelihood of subsequent events. For instance, if the first board is defective, the balance of defective vs. non-defective boards in the pool changes. Thus, picking the first board has a direct influence on the selection of the second, making them dependent. Understanding dependent events helps us gauge how prior outcomes can influence future predictions and adjustments in probability.
Conditional Probability
Conditional probability measures the likelihood of an event occurring given that another event has already occurred. It's expressed as \(P(A \mid B)\), meaning the probability of event \(A\) happening if \(B\) has already happened. In our case, we look at how the choice of the first circuit board affects the probability of the second one being defective. If you're already aware of the outcome of the first draw (whether or not it's defective), it influences the scenario of the second. So, \(P(E_2 \mid E_1)\) checks out the probability that the second board is defective given the first was, and \(P(E_2 \mid E_1^C)\) considers if the first wasn鈥檛. Conditional probabilities allow us to tailor our expectations based on conditions or outcomes already determined.
Independence of Events
Events are independent if the occurrence of one event does not affect the occurrence of another. For truly independent events, the probability of an event \(E_2\) should be the same, regardless of whether another event \(E_1\) has occurred. In terms of equation, \(P(E_2 \mid E_1) = P(E_2)\).
However, in reality, most events have some level of dependency. In the printed circuit board example, because selecting one board alters the pool from which you select the next, \(E_1\) and \(E_2\) start off as dependent. But, since the number of defective boards is so small compared to the total, their influence on the final outcomes is minimal, so they can be considered approximately independent in practical terms. Identifying independence helps simplify complex probability calculations by lessening the intricacy involved if dependencies are negligible.

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Most popular questions from this chapter

The National Public Radio show Car Talk has a feature called "The Puzzler." Listeners are asked to send in answers to some puzzling questions-usually about cars but sometimes about probability (which, of course, must account for the incredible popularity of the program!). Suppose that for a car question, 800 answers are submitted, of which 50 are correct. Suppose also that the hosts randomly select two answers from those submitted with replacement. a. Calculate the probability that both selected answers are correct. (For purposes of this problem, keep at least five digits to the right of the decimal.) b. Suppose now that the hosts select the answers at random but without replacement. Use conditional probability to evaluate the probability that both answers selected are correct. How does this probability compare to the one computed in Part (a)?

Consider the following information about travelers on vacation: \(40 \%\) check work email, \(30 \%\) use a cell phone to stay connected to work, \(25 \%\) bring a laptop with them on vacation, \(23 \%\) both check work email and use a cell phone to stay connected, and \(51 \%\) neither check work email nor use a cell phone to stay connected nor bring a laptop. In addition \(88 \%\) of those who bring a laptop also check work email 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 email, \(C=\) 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 \(L^{C}\) ) f. \(P(\) Eor C or \(L\) ) g. \(P(E \mid L)\) j. \(P(E\) and \(L)\) h. \(P(L \mid C)\) k. \(P(C\) and \(L)\) i. \(P(E\) and \(C\) and \(L)\) 1\. \(P(C \mid E\) and \(L)\)

The Associated Press (San Luis Obispo TelegramTribune, August 23,1995 ) reported on the results of mass screening of schoolchildren for tuberculosis (TB). For Santa Clara County, California, the proportion of all tested kindergartners who were found to have TB was .0006. The corresponding proportion for recent immigrants (thought to be a high-risk group) was .0075. Suppose that a Santa Clara County kindergartner is selected at random. Are the events selected student is a recent immigrant and selected student has \(T B\) independent or dependent events? Justify your answer using the given information.

A college library has four copies of a certain book; the copies are numbered \(1,2,3\), and 4 . Two of these are selected at random. The first selected book is placed on 2 -hr reserve, and the second book can be checked out overnight. a. Construct a tree diagram to display the 12 outcomes in the sample space. b. Let \(A\) denote the event that at least one of the books selected is an even-numbered copy. What outcomes are in \(A ?\) c. Suppose that copies 1 and 2 are first printings, whereas copies 3 and 4 are second printings. Let \(B\) denote the event that exactly one of the copies selected is a first printing. What outcomes are contained in \(B\) ?

A mutual fund company offers its customers several different funds: a money market fund, three different bond funds, two stock funds, and a balanced fund. Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows: \(\begin{array}{lr}\text { Money market } & 20 \% \\ \text { Short-term bond } & 15 \% \\ \text { Intermediate-term bond } & 10 \% \\ \text { Long-term bond } & 5 \% \\ \text { High-risk stock } & 18 \% \\ \text { Moderate-risk stock } & 25 \% \\ \text { Balanced fund } & 7 \%\end{array}\) A customer who owns shares in just one fund is to be selected at random. a. What is the probability that the selected individual owns shares in the balanced fund? b. What is the probability that the individual owns shares in a bond fund? c. What is the probability that the selected individual does not own shares in a stock fund?
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