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91Ó°ÊÓ

If \(P(B \mid A)>P(B)\), show that \(P\left(B^{\prime} \mid A\right)

Short Answer

Expert verified
If \(P(B | A)>P(B)\), then \(P(B' | A)<P(B')\).

Step by step solution

01

Understanding the Conditional Probability

To solve this problem, we need to understand the concept of conditional probability. The statement \( P(B | A) > P(B) \) means that the probability of event \( B \) occurring, given that \( A \) has occurred, is greater than the probability of event \( B \) occurring alone.
02

Using the Hint: Adding Probabilities

The hint suggests adding \( P(B' | A) \) to both sides of the inequality \( P(B | A) > P(B) \). This helps us utilize the law of total probability. Thus, we have:\[ P(B | A) + P(B' | A) > P(B) + P(B' | A) \] Since \( P(B | A) + P(B' | A) = 1 \), this simplifies to:\[ 1 > P(B) + P(B' | A) \].
03

Isolating the Probability

Rearrange the inequality from the previous step to isolate \( P(B' | A) \). This gives:\[ P(B' | A) < 1 - P(B) \].
04

Recognizing Complementarity

Recall that \( P(B') = 1 - P(B) \). Thus, the inequality \( P(B' | A) < 1 - P(B) \) can be rewritten as:\[ P(B' | A) < P(B') \]. This confirms the relationship between the conditional probability of the complement and its unconditional probability.

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Key Concepts

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

Law of Total Probability
The Law of Total Probability is a fundamental rule that helps us relate conditional probabilities with marginal probabilities. It provides a way to express the probability of an event, based on a partition of the sample space. Let's break this down simply:

If we have a set of mutually exclusive events \(A_1, A_2, \ldots, A_n\) that cover the entire sample space and an event \(B\), the probability of \(B\) can be expressed as:
  • \(P(B) = P(B \cap A_1) + P(B \cap A_2) + \ldots + P(B \cap A_n)\)
A more useful form is using conditional probabilities, which states:
  • \(P(B) = P(B \mid A_1)P(A_1) + P(B \mid A_2)P(A_2) + \ldots + P(B \mid A_n)P(A_n)\)
In the context of the given problem, we use the concept of conditional probabilities like \(P(B \mid A)\) and \(P(B' \mid A)\) to establish comparisons. Here, the law helps in connecting these probabilities by using the fact that their sum is always one: \(P(B \mid A) + P(B' \mid A) = 1\).

This understanding is key in analyzing how observing event \(A\) effects both \(B\) and its complement \(B'\).
Complementary Events
Complementary events are pairs of events where one event occurs if and only if the other does not. Mathematically, if \(B\) is an event, then its complement \(B'\) occurs whenever \(B\) does not. This means:
  • \(P(B) + P(B') = 1\)
In terms of probability, this relationship ensures that there is a balance. The total probability across the outcomes covered by both \(B\) and \(B'\) adds up to certainty (1).

In the problem you are working on, this principle is crucial to understand the inequality \(P(B' \mid A) < P(B')\). Knowing that the probability of an event and its complement sums to one helps reveal changes in the probability landscape due to additional information, like knowing event \(A\) has occurred.

This underpins the step where we rewrite \(1 - P(B)\) as \(P(B')\) and translate that into the inequality involving conditional probabilities. The complement's altered condition an observation must inform is how likelihoods adjust when prefacing information about \(A\) is engaged.
Probability Inequality
In the realm of probability, inequalities often arise when comparing conditional probability against unconditional probability. These can reveal relationships and insights about how events relate under different conditions.

In this specific problem, the inequality \(P(B \mid A) > P(B)\) indicates that knowing \(A\) has occurred increases the likelihood of event \(B\). This sets a context: event \(A\) positively correlates with \(B\). When using the complement \(B'\), the original inequality transforms to show that the chance of \(B'\) is actually reduced when \(A\) occurs, as summarized in \(P(B' \mid A) < P(B')\).
  • This highlights that the condition set by \(A\) skews probability favorably to \(B\)'s realization over its absence.
  • It helps conceptualize the events' dynamics, showcasing interdependencies and impacts conditional information can assert.
The inequality presents a powerful way to understand shifts in probability landscapes when additional conditions inform event likelihoods. By evaluating inequalities, one gains insight into how new information rearranges the probabilities of events occurring.

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Most popular questions from this chapter

Deer ticks can be carriers of either Lyme disease or human granulocytic ehrlichiosis (HGE). Based on a recent study, suppose that \(16 \%\) of all ticks in a certain location carry Lyme disease, \(10 \%\) carry HGE, and \(10 \%\) of the ticks that carry at least one of these diseases in fact carry both of them. If a randomly selected tick is found to have carried HGE, what is the probability that the selected tick is also a carrier of Lyme disease?

In five-card poker, a straight consists of five cards with adjacent denominations (e.g., 9 of clubs, 10 of hearts, jack of hearts, queen of spades, and king of clubs). Assuming that aces can be high or low, if you are dealt a five-card hand, what is the probability that it will be a straight with high card 10 ? What is the probability that it will be a straight? What is the probability that it will be a straight flush (all cards in the same suit)?

Four universities-1,2, 3, and 4 -are participating in a holiday basketball tournament. In the first round, 1 will play 2 and 3 will play 4 . Then the two winners will play for the championship, and the two losers will also play. One possible outcome can be denoted by 1324 ( 1 beats 2 and 3 beats 4 in first- round games, and then 1 beats 3 and 2 beats 4 ). a. List all outcomes in \(\mathcal{S}\). b. Let \(A\) denote the event that 1 wins the tournament. List outcomes in \(A\). c. Let \(B\) denote the event that 2 gets into the championship game. List outcomes in \(B\). d. What are the outcomes in \(A \cup B\) and in \(A \cap B\) ? What are the outcomes in \(A^{\prime}\) ?

An insurance company offers four different deductible levels-none, low, medium, and high-for its homeowner's policyholders and three different levels- low, medium, and high-for its automobile policyholders. The accompanying table gives proportions for the various categories of policyholders who have both types of insurance. For example, the proportion of individuals with both low homeowner's deductible and low auto deductible is \(.06\) (6\% of all such individuals). $$ \begin{array}{lcccc} && {\text { Homeowner's }} \\ \text { Auto } & \mathbf{N} & \mathbf{L} & \mathbf{M} & \mathbf{H} \\ \hline \mathbf{L} & .04 & .06 & .05 & .03 \\ \mathbf{M} & .07 & .10 & .20 & .10 \\ \mathbf{H} & .02 & .03 & .15 & .15 \\ \hline \end{array} $$ Suppose an individual having both types of policies is randomly selected. a. What is the probability that the individual has a medium auto deductible and a high homeowner's deductible? b. What is the probability that the individual has a low auto deductible? A low homeowner's deductible? c. What is the probability that the individual is in the same category for both auto and homeowner's deductibles? d. Based on your answer in part (c), what is the probability that the two categories are different? e. What is the probability that the individual has at least one low deductible level? f. Using the answer in part (e), what is the probability that neither deductible level is low?

The three most popular options on a certain type of new car are a built-in GPS \((A)\), a sunroof \((B)\), and an automatic transmission \((C)\). If \(40 \%\) of all purchasers request \(A, 55 \%\) request \(B, 70 \%\) request \(C, 63 \%\) request \(A\) or \(B\), \(77 \%\) request \(A\) or \(C, 80 \%\) request \(B\) or \(C\), and \(85 \%\) request \(A\) or \(B\) or \(C\), determine the probabilities of the following events. [Hint: " \(A\) or \(B\) " is the event that at least one of the two options is requested; try drawing a Venn diagram and labeling all regions.] a. The next purchaser will request at least one of the three options. b. The next purchaser will select none of the three options. c. The next purchaser will request only an automatic transmission and not either of the other two options. d. The next purchaser will select exactly one of these three options.

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