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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

Two different airlines have a flight from Los Angeles to New York that departs each weekday morning at a certain time. Let \(E\) denote the event that the first airline's flight is fully booked on a particular day, and let \(F\) denote the event that the second airline's flight is fully booked on that same day. Suppose that \(P(E)=.7, P(F)=.6\), and \(P(E \cap F)=.54\). a. Calculate \(P(E \mid F)\) the probability that the first airline's flight is fully booked given that the second airline's flight is fully booked. b. Calculate \(P(F \mid E)\).

A construction firm bids on two different contracts. Let \(E_{1}\) be the event that the bid on the first contract is successful, and define \(E_{2}\) analogously for the second contract. Suppose that \(P\left(E_{1}\right)=.4\) and \(P\left(E_{2}\right)=.2\) and that \(E_{1}\) and \(E_{2}\) are independent events. a. Calculate the probability that both bids are successful (the probability of the event \(E_{1}\) and \(E_{2}\) ). b. Calculate the probability that neither bid is successful (the probability of the event \(\left(\right.\) not \(\left.E_{1}\right)\) and \(\left(\right.\) not \(\left.E_{2}\right)\) ). c. What is the probability that the firm is successful in at least one of the two bids?

Suppose that, starting at a certain time, batteries coming off an assembly line are examined one by one to see whether they are defective (let \(\mathrm{D}=\) defective and \(\mathrm{N}=\) not defective). The chance experiment terminates as soon as a nondefective battery is obtained. a. Give five possible experimental outcomes. b. What can be said about the number of outcomes in the sample space? c. What outcomes are in the event \(E\), that the number of batteries examined is an even number?

Components of a certain type are shipped to a supplier in batches of \(10 .\) Suppose that \(50 \%\) of all batches contain no defective components, \(30 \%\) contain one defective component, and \(20 \%\) contain two defective components. A batch is selected at random. Two components from this batch are randomly selected and tested. a. If the batch from which the components were selected actually contains two defective components, what is the probability that neither of these is selected for testing? b. What is the probability that the batch contains two defective components and that neither of these is selected for testing? c. What is the probability that neither component selected for testing is defective? (Hint: This could happen with any one of the three types of batches. A tree diagram might help.)

A certain university has 10 vehicles available for use by faculty and staff. Six of these are vans and four are cars. On a particular day, only two requests for vehicles have been made. Suppose that the two vehicles to be assigned are chosen in a completely random fashion from among the 10 . a. Let \(E\) denote the event that the first vehicle assigned is a van. What is \(P(E)\) ? b. Let \(F\) denote the probability that the second vehicle assigned is a van. What is \(P(F \mid E)\) ? c. Use the results of Parts (a) and (b) to calculate \(P(E\) and \(F)\) (Hint: Use the definition of \(P(F \mid E) .)\)
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