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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 \(P\left(E_{2}\right)=.3\) 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 \(\left(\right.\) 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 .12\nb. The probability that neither bid is successful is .42\nc. The probability that the firm is successful in at least one of the bids is .58.

Step by step solution

01

Calculate the probability that both bids are successful

Using the definition of independent events, we can directly calculate the probability that both bids are successful as the product of the probabilities of each bid being successful: \(P(E_{1} \cap E_{2}) = P(E_{1})P(E_{2}) = .4*.3 = .12 \)
02

Calculate the probability that neither bid is successful

We can calculate the probability that each bid is unsuccessful as \(1-P(E_{i})\) where i is the bid number. Thus, the probability that neither bid is successful is the product of the probabilities that each bid is unsuccessful: \(P(\sim E_{1} \cap \sim E_{2}) = P(\sim E_{1})P(\sim E_{2}) = (1-.4)*(1-.3) = .42\)
03

Calculate the probability that the firm is successful in at least one of the bids

The probability that at least one bid is successful is the complement of the probability that neither bid is successful, or \(1 - P(\sim E_{1} \cap \sim E_{2}) = 1-.42 = .58\)

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Key Concepts

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

Independent Events
Independent events are a foundational concept in probability theory. Two events are considered independent when the occurrence of one does not affect the probability of the other happening.
A simple way to understand this is through everyday life examples. Think of flipping a coin and rolling a die simultaneously.
  • If the coin lands on heads, the result of the die roll remains unaffected.
  • Similarly, whether the die shows a three doesn't change the outcome of the coin flip.
In mathematical terms, for two events, such as our bids on contracts, to be independent, the probability of both occurring together is the product of their individual probabilities.
Thus, if event \(E_1\) has a probability \(P(E_1) = 0.4\) and event \(E_2\) has a probability \(P(E_2) = 0.3\), then the probability of both occurring, \(P(E_1 \cap E_2)\), is found by multiplying their probabilities: \(0.4 \times 0.3 = 0.12\).
Recognizing independent events quickly allows us to streamline probability calculations substantially.
Complementary Events
In probability, complementary events involve a situation where one event occurs, and its counterpart does not.
Essentially, these two events form a complete set of possible outcomes.
  • For example, if the probability of a rainy day is 0.2, the probability that it does not rain is 1 - 0.2 = 0.8.
  • This non-occurrence event is referred to as the complement.
This concept is especially useful when it's easier to calculate the probability of an event's complement rather than the event itself. In our contract scenario, determining the probability of neither bid being successful involves looking at the complements of each bid.
Hence, if the probability of a successful first bid is 0.4, its complement, i.e., an unsuccessful bid, becomes \(1 - 0.4 = 0.6\).
The same process applies to the second bid. Therefore, the probability of neither bid succeeding is \(0.6 \times 0.7 = 0.42\).
This use of complementary events can simplify even complex probability challenges.
Probability Calculations
Probability calculations often involve determining the likelihood of various outcomes. These calculations can typically be broken down based on key principles such as independent and complementary events.
  • The probability of multiple independent events occurring together is the product of their individual probabilities.
  • The probability of at least one event occurring is often calculated using the concept of complements.
For instance, in our exercise, the success of at least one bid is the complement of neither bid being successful. If the probability of neither bid succeeding is 0.42, the probability of at least one succeeding is calculated as \(1 - 0.42 = 0.58\).
These straightforward calculations simplify complex scenarios into manageable problems. Understanding these principles can aid significantly in interpreting and solving a wide range of probability-related questions.
With these foundational tools, students can tackle more intricate probability problems progressively.

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Most popular questions from this chapter

Three friends \((A, B\), and \(C)\) will participate in a round-robin tournament in which each one plays both of the others. Suppose that \(P(\mathrm{~A}\) beats \(\mathrm{B})=.7, P(\mathrm{~A}\) beats C) \(=.8, P(\mathrm{~B}\) beats \(\mathrm{C})=.6\), and that the outcomes of the three matches are independent of one another. a. What is the probability that \(\mathrm{A}\) wins both her matches and that \(\mathrm{B}\) beats \(\mathrm{C}\) ? b. What is the probability that \(A\) wins both her matches? c. What is the probability that \(A\) loses both her matches? d. What is the probability that each person wins one match? (Hint: There are two different ways for this to happen.)

A company uses three different assembly lines \(A_{1}, A_{2}\), and \(A_{3}\) -to manufacture a particular component. Of those manufactured by \(A_{1}, 5 \%\) need rework to remedy a defect, whereas \(8 \%\) of \(A_{2}\) 's components and \(10 \%\) of \(A_{3}\) 's components need rework. Suppose that \(50 \%\) of all components are produced by \(A_{1}\), whereas \(30 \%\) are produced by \(A_{2}\) and \(20 \%\) come from \(A_{3}\). a. Construct a tree diagram with first-generation branches corresponding to the three lines. Leading from each branch, draw one branch for rework (R) and another for no rework (N). Then enter appropriate probabilities on the branches. b. What is the probability that a randomly selected component came from \(A_{1}\) and needed rework? c. What is the probability that a randomly selected component needed rework?

The newspaper article "Folic Acid Might Reduce Risk of Down Syndrome" (USA Today, September 29 . 1999) makes the following statement: "Older women are at a greater risk of giving birth to a baby with Down Syndrome than are younger women. But younger women are more fertile, so most children with Down Syndrome are born to mothers under \(30 . "\) Let \(D=\) event that a randomly selected baby is born with Down Syndrome and \(Y=\) event that a randomly selected baby is born to a young mother (under age 30 ). For each of the following probability statements, indicate whether the statement is consistent with the quote from the article, and if not, explain why not. a. \(P(D \mid Y)=.001, P\left(D \mid Y^{C}\right)=.004, P(Y)=.7\) b. \(P(D \mid Y)=.001, P\left(D \mid Y^{C}\right)=.001, P(Y)=.7\) c. \(P(D \mid Y)=.004, P\left(D \mid Y^{C}\right)=.004, P(Y)=.7\) d. \(P(D \mid Y)=.001, P\left(D \mid Y^{C}\right)=.004, P(Y)=.4\) e. \(P(D \mid Y)=.001, P\left(D \mid Y^{C}\right)=.001, P(Y)=.4\) f. \(P(D \mid Y)=.004, P\left(D \mid Y^{C}\right)=.004, P(Y)=.4\)

Consider a system consisting of four components, as pictured in the following diagram: Components 1 and 2 form a series subsystem, as do Components 3 and 4 . The two subsystems are connected in parallel. Suppose that \(P(1\) works \()=.9, P(2\) works \()=\) \(.9, P(3\) works \()=.9\), and \(P(4\) works \()=.9\) and that the four components work independently of one another. a. The 1-2 subsystem works only if both components work. What is the probability of this happening? b. What is the probability that the \(1-2\) subsystem doesn't work? that the \(3-4\) subsystem doesn't work? c. The system won't work if the \(1-2\) subsystem doesn't work and if the \(3-4\) subsystem also doesn't work. What is the probability that the system won't work? that it will work? d. How would the probability of the system working change if a \(5-6\) subsystem were added in parallel with the other two subsystems? e. How would the probability that the system works change if there were three components in series in each of the two subsystems?

A large department store offers online ordering. When a purchase is made online, the customer can select one of four different delivery options: expedited overnight delivery, expedited second-business-day delivery, standard delivery, or delivery to the nearest store for customer pick-up. Consider the chance experiment that consists of observing the selected delivery option for a randomly selected online purchase. a. What are the simple events that make up the sample space for this experiment? b. Suppose that the probability of an overnight delivery selection is \(.1\), the probability of a second-day delivery selection is \(.3\), and the probability of a standard-delivery selection is . \(4 .\) Find the following probabilities: i. the probability that a randomly selected online purchase selects delivery to the nearest store for customer pick-up. ii. the probability that the customer selects a form of expedited delivery. iii. the probability that either standard delivery or delivery to the nearest store is selected.
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