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Chapter 5: Problem 64

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 \(P\left(E_{1}\right)=0.4\) and \(P\left(E_{2}\right)=0.3\) 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 \(\operatorname{not} E_{1}\) and not \(E_{2}\) ). 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 of winning both bids is 0.12. b. The probability of not winning any bids is 0.42. c. The probability of being successful in at least one of the two bids is 0.58.

Step by step solution

01

a. Probability of winning both bids

Since \(E_1\) and \(E_2\) are independent events, their joint probability can be calculated by simply multiplying their individual probabilities: \[P(E_1 \cap E_2) = P(E_1) \cdot P(E_2)\] Using the given probabilities: \[P(E_1 \cap E_2) = 0.4 \cdot 0.3 = 0.12\] The probability of winning both bids is 0.12.
02

b. Probability of not winning any bids

First, we need to find the probability of not winning each bid individually. This can be calculated as the complement: \[P(\text{not }E_1) = 1 - P(E_1) = 1 - 0.4 = 0.6\] \[P(\text{not }E_2) = 1 - P(E_2) = 1 - 0.3 = 0.7\] Now, we can find the probability of not winning both bids by multiplying the individual probabilities, since they are independent events: \[P(\text{not }E_1 \cap \text{not }E_2) = P(\text{not }E_1) \cdot P(\text{not }E_2) = 0.6 \cdot 0.7 = 0.42\] The probability of not winning any bids is 0.42.
03

c. Probability of winning at least one bid

To calculate this probability, we can use the complementary approach: find the probability of not winning both bids (which we calculated in part b), and subtract this value from 1: \[P(\text{at least one successful bid}) = 1 - P(\text{not }E_1 \cap \text{not }E_2)\] Using the result from part b: \[P(\text{at least one successful bid}) = 1 - 0.42 = 0.58\] The probability of being successful in at least one of the two bids is 0.58.

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Key Concepts

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

Independent Events
In probability, events are termed **independent** when the occurrence of one event does not affect the occurrence of another. For instance, when a construction firm bids on two separate contracts, the outcome of the bid on the first contract does not influence the outcome of the bid on the second. They operate independently.

When events are independent, the probability of both events occurring simultaneously can be found by multiplying the probabilities of each event. Mathematically, for two events, \(E_{1}\) and \(E_{2}\), we have:
  • \(P(E_{1} \text{ and } E_{2}) = P(E_{1}) \times P(E_{2})\)
This concept simplifies computation because there is no need to consider additional factors.

In the given exercise, the probability of both bids being successful is determined by multiplying the probabilities of individual successes.
Complementary Probability
The **complement** of an event is essentially the probability of the event not occurring. This concept is crucial because it allows us to calculate probabilities indirectly.

If an event \(E\) has a probability \(P(E)\), then the probability of the complement of \(E\), denoted \(\text{not } E\), is given by:
  • \(P(\text{not } E) = 1 - P(E)\)
This subtraction from one accounts for the fact that the total probability of all possible outcomes in any scenario is always equal to one.

In our exercise, this concept is used to find the probability of not securing either contract bid. By finding the complement probabilities of winning each bid, we then calculate the probability of not winning any bids at all.
Joint Probability
**Joint probability** refers to the likelihood of two events occurring at the same time. When the events are independent, this probability can be computed by multiplying their individual probabilities. Hence, the joint probability of events \(E_{1}\) and \(E_{2}\) is depicted as \(P(E_{1} \cap E_{2})\).

However, if events are dependent, the joint probability would require considering the conditional probability affected by one event over the other, a scenario not applicable in this exercise.

In the context of our problem, where the construction firm's success on both bids is concerned, the joint probability \(P(E_{1} \cap E_{2})\) illustrates the likelihood of both outcomes occurring together. This is computed straightforwardly due to the independence of the events involved.
Event Probability
**Event probability** is the fundamental measure indicating the likelihood of a particular outcome occurring. It's expressed as a value between 0 and 1, where 0 implies impossibility, and 1 indicates certainty.

For any event \(E\), \(P(E)\) quantifies its probability, serving as a building block for more complex probability calculations, such as joint or complementary probabilities.

In this exercise, the probabilities of each bidding success \(P(E_{1})\) and \(P(E_{2})\) are given directly. These provide essential starting points for computing the joint probability of success on both bids, as well as the result of complementary probabilities to determine the chance of at least one bid failing.

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Most popular questions from this chapter

The following table summarizes data on smoking status and age group, and is consistent with summary quantities obtained in a Gallup Poll published in the online article "In U.S., Young Adults' Cigarette Use Is Down Sharply" $$ \begin{array}{|lcc|} \hline & {\text { Smoking Status }} \\ \hline { 2 - 3 } \text { Age Group } & \text { Smoker } & \text { Nonsmoker } \\\ \hline 18 \text { to } 29 & 174 & 618 \\ 30 \text { to } 49 & 333 & 1,115 \\ 50 \text { to } 64 & 384 & 1,445 \\ 65 \text { and older } & 211 & 1,707 \\ \hline \end{array} $$ Assume that it is reasonable to consider these data as representative of the American adult population. Consider the chance experiment or randomly selecting an adult American. a. What is the probability that the selected adult is a smoker? b. What is the probability that the selected adult is under 50 years of age? c. What is the probability that the selected adult is a smoker that is 65 or older? d. What is the probability that the selected adult is a smoker or is age 65 or older?

A large cable company reports that \(80 \%\) of its customers subscribe to its cable TV service, \(42 \%\) subscribe to its Internet service, and \(97 \%\) subscribe to at least one of these two services. a. Use the given probability information to set up a hypothetical 1000 table. b. Use the table from Part (a) to find the following probabilities: i. the probability that a randomly selected customer subscribes to both cable TV and Internet service. ii. the probability that a randomly selected customer subscribes to exactly one of these services.

A large cable company reports the following: \(80 \%\) of its customers subscribe to cable TV service \(42 \%\) of its customers subscribe to Internet service \(32 \%\) of its customers subscribe to telephone service \(25 \%\) of its customers subscribe to both cable TV and Internet service \(21 \%\) of its customers subscribe to both cable TV and phone service \(23 \%\) of its customers subscribe to both Internet and phone service \(15 \%\) of its customers subscribe to all three services Consider the chance experiment that consists of selecting one of the cable company customers at random. In Exercise \(5.53,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. \(P(\) cable TV only) b. \(P\) (Internet \(\mid\) cable TV) c. \(P(\) exactly two services \()\) d. \(P\) (Internet and cable TV only)

5.57 There are two traffic lights on Shelly's route from home to work. Let \(E\) denote the event that Shelly must stop at the first light, and define the event \(F\) in a similar manner for the second light. Suppose that \(P(E)=0.4, P(F)=0.3\) and \(P(E \cap F)=0.15 .\) In Exercise \(5.25,\) you constructed a hypothetical 1000 table to calculate the following probabilities. Now use the probability formulas of this section to find these probabilities. a. The probability that Shelly must stop for at least one light (the probability of the event \(E \cup F\) ). b. The probability that Shelly does not have to stop at either light. c. The probability that Shelly must stop at exactly one of the two lights. d. The probability that Shelly must stop only at the first light.

A single-elimination tournament with four players is to be held. A total of three games will be played. In Game 1 , the players seeded (rated) first and fourth play. In Game 2 , the players seeded second and third play. In Game \(3,\) the winners of Games 1 and 2 play, with the winner of Game 3 declared the tournament winner. Suppose that the following probabilities are known: $$ P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 4)=0.8 $$ \(P(\) Seed 1 defeats \(\operatorname{Seed} 2)=0.6\) $$ P(\text { Seed } 1 \text { defeats } \operatorname{Seed} 3)=0.7 $$ \(P(\) Seed 2 defeats Seed 3\()=0.6\) \(P(\) Seed 2 defeats Seed 4\()=0.7\) \(P(\) Seed 3 defeats \(\operatorname{Seed} 4)=0.6\) a. How would you use random digits to simulate Game 1 of this tournament? b. How would you use random digits to simulate Game 2 of this tournament? c. How would you use random digits to simulate the third game in the tournament? (This will depend on the outcomes of Games 1 and \(2 .\) ) d. Simulate one complete tournament, giving an explanation for each step in the process. e. Simulate 10 tournaments, and use the resulting information to estimate the probability that the first seed wins the tournament. f. Ask four classmates for their simulation results. Along with your own results, this should give you information on 50 simulated tournaments. Use this information to estimate the probability that the first seed wins the tournament. g. Why do the estimated probabilities from Parts (e) and (f) differ? Which do you think is a better estimate of the actual probability? Explain.
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