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If \(A\) and \(B\) are independent events, show that \(A^{\prime}\) and \(B\) are also independent. [Hint: First establish a relationship between \(P\left(A^{\prime} \cap B\right), P(B)\), and \(P(A \cap B)\).]

Short Answer

Expert verified
\(A'\) and \(B\) are independent because \(P(A' \cap B) = P(B) \times P(A')\).

Step by step solution

01

Understanding Independence

For two events \(A\) and \(B\) to be independent, the probability of their intersection equals the product of their individual probabilities. Thus, \(A\) and \(B\) are independent if \(P(A \cap B) = P(A) \times P(B)\).
02

Express Complement of Event

The complement of an event \(A\) is denoted as \(A'\), which represents all outcomes not in \(A\). The probability of the complement is \(P(A') = 1 - P(A)\).
03

Intersection of Complement and Event

We need to find \(P(A' \cap B)\). Using the property \(P(A' \cap B) = P(B) - P(A \cap B)\), since \(A' \cap B\) includes those outcomes in \(B\) but not in \(A\).
04

Substitute the Independence Condition

Substituting \(P(A \cap B) = P(A) \times P(B)\) into \(P(A' \cap B) = P(B) - P(A \cap B)\), we get \(P(A' \cap B) = P(B) - P(A) \times P(B)\).
05

Factor and Simplify

Factor \(P(B)\) out of the expression to obtain \(P(A' \cap B) = P(B) \times (1 - P(A))\). This can be rewritten as \(P(A' \cap B) = P(B) \times P(A')\).
06

Confirm Independence of Complement and Event

Since \(P(A' \cap B) = P(B) \times P(A')\), this confirms that \(A'\) and \(B\) are independent. This follows the definition that two events \(X\) and \(Y\) are independent if and only if \(P(X \cap Y) = P(X) \times P(Y)\).

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Key Concepts

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

Probability of Intersection
When studying probability, understanding the probability of intersection between two events is crucial. This refers to the likelihood of both events happening at the same time. For two events, say \(A\) and \(B\), the intersection is denoted by \(A \cap B\), meaning that both events occur.
The formula for the probability of the intersection of two independent events \(A\) and \(B\) is given by \(P(A \cap B) = P(A) \times P(B)\).
  • This formula works when the occurrence of one event doesn't affect the other.
  • It's important because it helps to assess how likely two things are to happen simultaneously.
For example, if you have a red die and a blue die, the chance of rolling number 4 on both dies is the intersection of rolling a 4 on the red die and rolling a 4 on the blue die. If both are fair six-sided dies, the probability would be \(\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}\).
Understanding this concept allows us to tackle more complex problems in probability involving multiple events and their interactions.
Complement of an Event
In probability, the complement of an event is everything that is not in the event. If an event \(A\) represents a successful outcome, then \(A'\), called "A complement," includes all other outcomes.
We use the expression \(P(A') = 1 - P(A)\) to calculate the likelihood that event \(A\) does not occur.
  • This is a powerful tool for solving probability problems, providing a straightforward method to find the probability of the opposite of an event.
  • Complements are used when it's easier to calculate the probability of an event not happening.
For instance, if you have a 90% chance of raining (event \(A\)), then there's a 10% chance that it will not rain (event \(A'\)). Understanding complements aids in visualizing all possible outcomes in probability and helps in calculating probabilities of complex actions by simplifying into straightforward complementary actions.
Event Independence
Event independence is a fundamental concept in probability that defines specific relationships between events. Two events \(A\) and \(B\) are considered independent if the occurrence of one does not affect the occurrence of another.
Mathematically, \(A\) and \(B\) are independent if \(P(A \cap B) = P(A) \times P(B)\). This implies that knowing whether \(A\) happens does not provide any information about \(B\).
  • Independence allows simplifying complex probability problems because we can multiply the probabilities of individual events rather than calculate them altogether.
  • This concept underpins many statistical principles and is crucial for understanding real-world scenarios where one event's outcome is entirely unaffected by another.
For instance, the flip of a coin is independent of the roll of a die. The result of getting heads or tails does not alter the outcome of a die roll. Recognizing independence in problems can make tough problems easier to tackle, as it allows for straightforward computations and understanding of how events interact.

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Most popular questions from this chapter

A quality control inspector is inspecting newly produced items for faults. The inspector searches an item for faults in a series of independent fixations, each of a fixed duration. Given that a flaw is actually present, let \(p\) denote the probability that the flaw is detected during any one fixation (this model is discussed in "Human Performance in Sampling Inspection," Human Factors, 1979: 99-105). a. Assuming that an item has a flaw, what is the probability that it is detected by the end of the second fixation (once a flaw has been detected, the sequence of fixations terminates)? b. Give an expression for the probability that a flaw will be detected by the end of the \(n\)th fixation. c. If when a flaw has not been detected in three fixations, the item is passed, what is the probability that a flawed item will pass inspection? d. Suppose \(10 \%\) of all items contain a flaw \([P\) (randomly chosen item is flawed) \(=.1]\). With the assumption of part (c), what is the probability that a randomly chosen item will pass inspection (it will automatically pass if it is not flawed, but could also pass if it is flawed)? e. Given that an item has passed inspection (no flaws in three fixations), what is the probability that it is actually flawed? Calculate for \(p=.5\).

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 stereo store is offering a special price on a complete set of components (receiver, compact disc player, speakers, cassette deck). A purchaser is offered a choice of manufacturer for each component: Receiver: Kenwood, Onkyo, Pioneer, Sony, Sherwood Compact disc player: Onkyo, Pioneer, Sony, Technics Speakers: Boston, Infinity, Polk Cassette deck: Onkyo, Sony, Teac, Technics A switchboard display in the store allows a customer to hook together any selection of components (consisting of one of each type). Use the product rules to answer the following questions: a. In how many ways can one component of each type be selected? b. In how many ways can components be selected if both the receiver and the compact disc player are to be Sony? c. In how many ways can components be selected if none is to be Sony? d. In how many ways can a selection be made if at least one Sony component is to be included? e. If someone flips switches on the selection in a completely random fashion, what is the probability that the system selected contains at least one Sony component? Exactly one Sony component?

The probability that a grader will make a marking error on any particular question of a multiple-choice exam is .1. If there are ten questions and questions are marked independently, what is the probability that no errors are made? That at least one error is made? If there are \(n\) questions and the probability of a marking error is \(p\) rather than \(.1\), give expressions for these two probabilities.

1,30 \%\( of the time on airline \)\… # A friend who lives in Los Angeles makes frequent consulting trips to Washington, D.C.; \(50 \%\) of the time she travels on airline \(\\# 1,30 \%\) of the time on airline \(\\# 2\), and the remaining \(20 \%\) of the time on airline #3. For airline #1, flights are late into D.C. \(30 \%\) of the time and late into L.A. \(10 \%\) of the time. For airline \(\\# 2\), these percentages are \(25 \%\) and \(20 \%\), whereas for airline \(\\# 3\) the percentages are \(40 \%\) and \(25 \%\). If we learn that on a particular trip she arrived late at exactly one of the two destinations, what are the posterior probabilities of having flown on airlines #1, #2, and #3? Assume that the chance of a late arrival in L.A. is unaffected by what happens on the flight to D.C. [Hint: From the tip of each first-generation branch on a tree diagram, draw three second-generation branches labeled, respectively, 0 late, 1 late, and 2 late.]

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