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Chapter 2: Problem 73

If \(A\) and \(B\) are independent events, show that \(A^{\prime}\) and \(B\) are also independent.

Short Answer

Expert verified
Events \(A^{\prime}\) and \(B\) are independent because \( P(A^{\prime} \cap B) = P(A^{\prime}) \times P(B) \).

Step by step solution

01

Review the Definition of Independence

Two events, say events \(A\) and \(B\), are said to be independent if the probability of both events occurring together is the product of their individual probabilities. This is mathematically stated as: \( P(A \cap B) = P(A) \times P(B) \).
02

Use Independence of Events A and B

Since events \(A\) and \(B\) are independent, we know that \( P(A \cap B) = P(A) \times P(B) \). This will be applied to simplify terms for \(A^{\prime}\), the complement of \(A\).
03

Express Complement Event Probability Relation

The probability of the complement of an event \(A\), which is \(A^{\prime}\), is given by \( P(A^{\prime}) = 1 - P(A) \). We will also use the fact that \( P(A^{\prime} \cap B) = P(B) - P(A \cap B) \).
04

Substitute the Known Values

Using the values from previous steps, substitute \( P(A^{\prime}) \) and \( P(B) \) into the relation: \( P(A^{\prime} \cap B) = P(B) - P(A \cap B) = P(B) - P(A) \times P(B) \).
05

Factorize the Expression

Factor \(P(B)\) from the modified relation: \( P(A^{\prime} \cap B) = P(B) (1 - P(A)) \).
06

Express Independence of A' and B

Observe that \( 1 - P(A) = P(A^{\prime}) \). Thus, we have: \( P(A^{\prime} \cap B) = P(B) \times P(A^{\prime}) \).
07

Conclude A' and B are Independent

Conclude that since \( P(A^{\prime} \cap B) \) equals \( P(A^{\prime}) \times P(B) \), the events \( A^{\prime} \) and \( B \) are independent.

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Key Concepts

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

Independent Events
In probability theory, independent events are crucial for understanding probabilities. Two events are considered independent if the occurrence of one does not affect the probability of the other. This means that understanding or knowing whether one event happened gives us no information about the other. To determine if two events, say event \( A \) and event \( B \), are independent, we can use their probabilities. If the combined probability of both events happening together, \( P(A \cap B) \), is the same as the product of their individual probabilities, \( P(A) \times P(B) \), then these events are independent. This relationship allows for a simpler calculation of complex probabilities by multiplying straightforward probabilities from each separate event. It鈥檚 like figuring out if two friends decided to show up at a party on their own without knowing or needing to check if the other was coming.
Complementary Events
Complementary events help simplify probability calculations. The complement of an event \( A \), written as \( A^{\prime} \), consists of all outcomes not in \( A \). For example, if event \( A \) is getting a heads in a coin toss, then \( A^{\prime} \) would be getting a tails. The sum of the probabilities of an event and its complement is always 1, meaning \( P(A) + P(A^{\prime}) = 1 \). This relationship allows us to easily find the probability of the complement once we know the probability of the event itself. The complement rule is handy, especially when determining probabilities of events not directly outlined or when the complement is easier to evaluate. Knowing \( P(A^{\prime}) \) as \( 1 - P(A) \) sets the stage for solving problems that deal with non-occurrence scenarios.
Probability of Events
The probability of events discusses how likely it is for any event to occur. It is essentially the measure of the chance that a given event will happen, quantified as a number between 0 and 1. A probability of 0 indicates impossibility, whereas a probability of 1 signifies certainty. Calculating the probability of events involves looking at all the possible outcomes and determining how many of these meet the criteria of the event. The basic formula to find the probability of an event \( A \) is \( P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \). By applying this formula, along with understanding independent and complementary events, students can break down complex probability problems into more straightforward parts. This methodical approach allows for a comprehensive view of probability, empowering students to solve and analyze different scenarios relatably and accurately.

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Most popular questions from this chapter

For customers purchasing a refrigerator at a certain appliance store, let \(A\) be the event that the refrigerator was manufactured in the U.S., \(B\) be the event that the refrigerator had an icemaker, and \(C\) be the event that the customer purchased an extended warranty. Relevant probabilities are $$ \begin{aligned} &P(A)=.75 \quad P(B \mid A)=.9 \quad P\left(B \mid A^{\prime}\right)=.8 \\ &P(C \mid A \cap B)=.8 \quad P\left(C \mid A \cap B^{\prime}\right)=.6 \\ &P\left(C \mid A^{\prime} \cap B\right)=.7 \quad P\left(C \mid A^{\prime} \cap B^{\prime}\right)=.3 \end{aligned} $$ a. Construct a tree diagram consisting of first-, second-, and third- generation branches, and place an event label and appropriate probability next to each branch. b. Compute \(P(A \cap B \cap C)\). c. Compute \(P(B \cap C)\). d. Compute \(P(C)\). e. Compute \(P(A \mid B \cap C)\), the probability of a U.S. purchase given that an icemaker and extended warranty are also purchased.

A transmitter is sending a message by 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 will be 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 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.] c. Suppose \(70 \%\) of all bits sent from the transmitter are \(1 \mathrm{~s}\). If a 1 is received by the receiver, what is the probability that a 1 was sent?

A small manufacturing company will start operating a night shift. There are 20 machinists employed by the company. a. If a night crew consists of 3 machinists, how many different crews are possible? b. If the machinists are ranked \(1,2, \ldots, 20\) in order of competence, how many of these crews would not have the best machinist? c. How many of the crews would have at least 1 of the 10 best machinists? d. If one of these crews is selected at random to work on a particular night, what is the probability that the best machinist will not work that night?

Each of a sample of four home mortgages is classified as fixed rate \((F)\) or variable rate \((V)\). a. What are the 16 outcomes in \(\mathcal{S}\) ? b. Which outcomes are in the event that exactly three of the selected mortgages are fixed rate? c. Which outcomes are in the event that all four mortgages are of the same type? d. Which outcomes are in the event that at most one of the four is a variable- rate mortgage? e. What is the union of the events in parts (c) and (d), and what is the intersection of these two events? f. What are the union and intersection of the two events in parts (b) and (c)?

Four engineers, \(A, B, C\), and \(D\), have been scheduled for job interviews at \(10 \mathrm{~A} . \mathrm{M}\). on Friday, January 13 , at Random Sampling, Inc. The personnel manager has scheduled the four for interview rooms \(1,2,3\), and 4 , respectively. However, the manager's secretary does not know this, so assigns them to the four rooms in a completely random fashion (what else!). What is the probability that a. All four end up in the correct rooms? b. None of the four ends up in the correct room?
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