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For any events \(A\) and \(B\) with \(P(B)>0\), show that \(P(A \mid B)+\) \(P\left(A^{\prime} \mid B\right)=1\).

Short Answer

Expert verified
The sum of conditional probabilities \(P(A \mid B) + P(A' \mid B)\) equals 1.

Step by step solution

01

Understand Conditional Probability

The conditional probability of event \(A\) given event \(B\) is defined as \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\). Similarly, \(P\left(A' \mid B\right)\) represents the probability of the complement of \(A\) given \(B\) and is defined as \(P\left(A' \mid B\right) = \frac{P(A' \cap B)}{P(B)}\).
02

Use Complement Rule in Probability

The complement rule for probabilities states that for any event \(A\), \(P(A) + P(A') = 1\). This rule also applies to conditional probabilities.
03

Apply Complement Rule to Conditional Probabilities

Using the complement rule in the context of conditional probabilities, we have \(P(A \mid B) + P(A' \mid B)\) to check if this sum equals 1. From step 2, we know \(P(A \cap B) + P(A' \cap B) = P(B)\) since \(A \cap B\) and \(A' \cap B\) are mutually exclusive and exhaustive within \(B\).
04

Simplify the Equation

Substitute \(P(A \cap B) + P(A' \cap B) = P(B)\) into the conditional probability expressions: \(\frac{P(A \cap B)}{P(B)} + \frac{P(A' \cap B)}{P(B)} = \frac{P(A \cap B) + P(A' \cap B)}{P(B)}\).
05

Calculate and Verify Result

Since \(P(A \cap B) + P(A' \cap B) = P(B)\), substituting gives \(\frac{P(B)}{P(B)} = 1\). Therefore, \(P(A \mid B) + P(A' \mid B) = 1\), confirming the validity of the equation.

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Key Concepts

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

Complement Rule
The complement rule is a fundamental concept in probability that simplifies the understanding of probabilities involving an event and its opposite. When dealing with an event \(A\), the complement of \(A\), often denoted as \(A'\), represents all outcomes not in \(A\). The probability of \(A\) and its complement always adds up to 1. This rule can be expressed as:
  • \(P(A) + P(A') = 1\)
Applying this rule to conditional probabilities involves some adjustments, but the principle holds. For any condition \(B\), the conditional probability of \(A\) and its complement given \(B\) is:
  • \(P(A \mid B) + P(A' \mid B) = 1\)
This formula shows that under a certain condition \(B\), either \(A\) happens or it doesn't, leaving no other possibilities within the framework established by \(B\). The complement rule allows for a straightforward method to resolve and verify probability calculations related to conditional events and their opposites.
Mutually Exclusive Events
Mutually exclusive events are a concept in probability that refers to two events that cannot occur simultaneously. In other words, the occurrence of one event excludes the occurrence of the other. For example, when flipping a coin, getting heads and getting tails are mutually exclusive events because both cannot occur in a single coin flip.
When examining two mutually exclusive events \(A\) and \(B\), the probability of both occurring at the same time is zero:
  • \(P(A \cap B) = 0\)
This notion is particularly useful when calculating probabilities within a condition. For instance, when dealing with conditional events \(A \cap B\) and \(A' \cap B\), they are mutually exclusive in the context of the broader condition \(B\). Thus, when calculating conditional probabilities, their sum simplifies to the probability of \(B\), which is crucial in showing that \(P(A \mid B) + P(A' \mid B) = 1\). Understanding mutual exclusivity helps in correctly applying probability rules to complex problems.
Exhaustive Events
Exhaustive events cover all possible outcomes of a probability experiment within a given sample space. If events are exhaustive, they collectively include every possible result that can occur in the experiment. For instance, tossing a six-sided die, the events of rolling 1, 2, 3, 4, 5, or 6 are exhaustive as they encompass the entire space of outcomes that can occur.
When considering any event \(A\) and its complement \(A'\), these two together form exhaustive events because they account for all outcomes, either \(A\) happens or \(A'\) happens. This completeness is expressed in:
  • \(P(A \cup A') = 1\)
In conditional probability, when examining events within a set condition \(B\), \(A \cap B\) and \(A' \cap B\) are exhaustive within \(B\).
Together, they cover all possible scenarios allowed by \(B\), ensuring no outcome outside \(B\) is overlooked. Recognizing events as exhaustive allows for the application of probability rules with confidence, ensuring complete analysis within the defined parameters.

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Most popular questions from this chapter

An academic department with five faculty membersAnderson, Box, Cox, Cramer, and Fisher-must select two of its members to serve on a personnel review committee. Because the work will be time-consuming, no one is anxious to serve, so it is decided that the representative will be selected by putting five slips of paper in a box, mixing them, and selecting two. a. What is the probability that both Anderson and Box will be selected? [Hint: List the equally likely outcomes.] b. What is the probability that at least one of the two members whose name begins with \(C\) is selected? c. If the five faculty members have taught for \(3,6,7,10\), and 14 years, respectively, at the university, what is the probability that the two chosen representatives have at least 15 years' teaching experience at the university?

An academic department with five faculty members narrowed its choice for department head to either candidate \(A\) or candidate \(B\). Each member then voted on a slip of paper for one of the candidates. Suppose there are actually three votes for \(A\) and two for \(B\). If the slips are selected for tallying in random order, what is the probability that \(A\) remains ahead of \(B\) throughout the vote count (e.g., this event occurs if the selected ordering is \(A A B A B\), but not for \(A B B A A)\) ?

Suppose identical tags are placed on both the left ear and the right ear of a fox. The fox is then let loose for a period of time. Consider the two events \(C_{1}=\) [left ear tag is lost \(\\}\) and \(C_{2}=\) [right ear tag is lost \(]\). Let \(\pi=P\left(C_{1}\right)=P\left(C_{2}\right)\), and assume \(C_{1}\) and \(C_{2}\) are independent events. Derive an expression (involving \(\pi\) ) for the probability that exactly one tag is lost given that at most one is lost ("Ear Tag Loss in Red Foxes," J. Wildife Mgmt., 1976: 164-167). [Hint: Draw a tree diagram in which the two initial branches refer to whether the left ear tag was lost.]

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?

Let \(A\) denote the event that the next request for assistance from a statistical software consultant relates to the SPSS package, and let \(B\) be the event that the next request is for help with SAS. Suppose that \(P(A)=.30\) and \(P(B)=.50\). a. Why is it not the case that \(P(A)+P(B)=1\) ? b. Calculate \(P\left(A^{\prime}\right)\). c. Calculate \(P(A \cup B)\). d. Calculate \(P\left(A^{\prime} \cap B^{\prime}\right)\).

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