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Chapter 6: Problem 60

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?

Short Answer

Expert verified
a. The probability that both bids are successful is 0.08. b. The probability that neither bid is successful is 0.48. c. The probability that at least one bid is successful is 0.52.

Step by step solution

01

Calculating the Probability that Both Bids are Successful

Given the two events \(E_{1}\) and \(E_{2}\) are independent, the probability of both events occurring simultaneously is the product of their individual probabilities. Therefore, the probability that both bids are successful, \(P(E_{1} \cap E_{2})\), can be calculated using the formula \(P(E_{1}) \cdot P(E_{2})\). From the problem, \(P(E_{1}) = 0.4\) and \(P(E_{2}) = 0.2\). So, \(P(E_{1} \cap E_{2}) = 0.4 \cdot 0.2 = 0.08\).
02

Calculating the Probability that Neither Bids is Successful

The probability of the complement of an event can be calculated as \(1 - P(E)\). Given that events \(E_{1}\) and \(E_{2}\) are independent, the probability that neither event occurs (not \(E_{1}\) and not \(E_{2}\)) is the product of their individual complement probabilities. The calculation would be as follows: \(P(\neg E_{1}) = 1 - P(E_{1}) = 1 - 0.4 = 0.6, P(\neg E_{2}) = 1 - P(E_{2}) = 1 - 0.2 = 0.8\), so \(P(\neg E_{1} \cap \neg E_{2}) = P(\neg E_{1}) \cdot P(\neg E_{2}) = 0.6 \cdot 0.8 = 0.48\).
03

Calculating the Probability that at Least One Bid is Successful

The event that at least one bid is successful is the complement of the event that neither bid is successful. So, its probability can be calculated as \(P(\left(E_{1} \cup E_{2}\right) = 1 - P(\neg E_{1} \cap \neg E_{2}) = 1 - 0.48 = 0.52\).

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Key Concepts

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

Statistical Independence
Understanding the concept of statistical independence is crucial when analyzing the relationship between two events. In probability theory, two events are considered to be statistically independent if the occurrence of one event has no effect on the probability of the occurrence of the other. To put it simply, knowing that one event has occurred does not change the likelihood of the other event happening.

For example, in the exercise provided, the fact that the construction firm is successful in one contract bid does not influence the outcome of the other bid. The events are independent, and their probabilities can be multiplied to find the combined probability of both events occurring. Formally, this relationship is expressed as:
\[ P(E_1 \cap E_2) = P(E_1) \cdot P(E_2) \].

It is important to note that independence is a theoretical assumption and may not always reflect real-world situations. Furthermore, the probability of independent events never adds up to more than 1, as they are separate occurrences not affecting each other.
Complement of an Event
The complement of an event is a fundamental concept in probability and represents all outcomes that are not part of the event itself. Formally, the complement of an event \(E\) is denoted as \(eg E\) and essentially captures the notion of 'everything else that can happen in our sample space outside of event \(E\)'.

If you're told that the probability of a certain event is, say, 0.4, then you can instantly determine that the probability of the event not happening is 0.6, because \(P(eg E) = 1 - P(E)\). This is precisely what we did in the textbook exercise for each bid. The calculation of the probability that neither bid is successful is also a direct application of this concept, where you multiply the probabilities of the complements of each independent event.

Another useful application of this concept is that it can help you determine the probability of at least one event occurring. By finding the probability that neither event occurs and then taking the complement of that, you're left with the probability that at least one event occurs, which is what part (c) of the exercise asked for.
Union of Events
The union of two events, denoted as \(E_1 \cup E_2\), represents the occurrence of at least one of the events. It's the event that either \(E_1\), \(E_2\), or both take place. In simpler terms, it's like saying 'either this or that or both.' This is particularly important for compound events where multiple outcomes are favorable.

Calculating the probability of the union of independent events requires careful attention. For mutually exclusive events (events that cannot happen at the same time), the probability of their union is the sum of their individual probabilities. However, when events are not mutually exclusive, as in they can both happen together, we must correct for the fact that we have counted the intersection twice. This correction leads us to the principle of inclusion-exclusion:
\[ P(E_1 \cup E_2) = P(E_1) + P(E_2) - P(E_1 \cap E_2) \].

In the context of the exercise, we were looking for the probability of at least one contract being successful. Since we learned that neither contract affects the other (independence), this can be simplified by finding the probability that both contracts fail and subtracting from one. Using the concept of the complement, this yields the probability of the union without having to add individual probabilities and subtract the intersection.

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Most popular questions from this chapter

Refer to Exercise 6.18. Adding probabilities in the first row of the given table yields \(P(\) midsize \()=.45\), whereas from the first column, \(\mathrm{P}\left(4 \frac{3}{8}\right.\) in. grip) \(=.30\). Is the following true? $$ P\left(\text { midsize } \text { or } 4 \frac{3}{8} \text { in. grip }\right)=.45+.30=.75 $$ Explain.

Only \(0.1 \%\) of the individuals in a certain population have a particular disease (an incidence rate of .001). Of those who have the disease, \(95 \%\) test positive when a certain diagnostic test is applied. Of those who do not have the disease, \(90 \%\) test negative when the test is applied. Suppose that an individual from this population is randomly selected and given the test. a. Construct a tree diagram having two first-generation branches, for has disease and doesn't have disease, and two second-generation branches leading out from each of these, for positive test and negative test. Then enter appropriate probabilities on the four branches. b. Use the general multiplication rule to calculate \(P(\) has disease and positive test). c. Calculate \(P\) (positive test). d. Calculate \(P\) (has disease \(\mid\) positive test). Does the result surprise you? Give an intuitive explanation for the size of this probability.

A transmitter is sending a message using a binary code, namely, a sequence of 0's and 1's. Each transmitted bit \((0\) or 1\()\) must pass through three relays to reach the receiver. At each relay, the probability is \(.20\) that the bit sent on is different from the bit received (a reversal). Assume that the relays operate independently of one another: transmitter \(\rightarrow\) relay \(1 \rightarrow\) relay \(2 \rightarrow\) relay \(3 \rightarrow\) receiver a. If a 1 is sent from the transmitter, what is the probability that a 1 is sent on by all three relays? b. If a 1 is sent from the transmitter, what is the probability that a 1 is received by the receiver? (Hint: The eight experimental outcomes can be displayed on a tree diagram with three generations of branches, one generation for each relay.)

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 \(P(F \mid E)\).

According to a study released by the federal Substance Abuse and Mental Health Services Administration (Knight Ridder Tribune, September 9,1999 ), approximately \(8 \%\) of all adult full-time workers are drug users and approximately \(70 \%\) of adult drug users are employed full-time. a. Is it possible for both of the reported percentages to be correct? Explain. b. Define the events \(D\) and \(E\) as \(D=\) event that a randomly selected adult is a drug user and \(E=\) event that a randomly selected adult is employed full- time. What are the estimated values of \(P(D \mid E)\) and \(P(E \mid D) ?\) c. Is it possible to determine \(P(D)\), the probability that a randomly selected adult is a drug user, from the information given? If not, what additional information would be needed?
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