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Prove that if \(P(B \mid A)>P(B)\) [in which case we say that "A attracts \(B\) "], then \(P(A \mid B)>P(A)\) [" \(B\) attracts \(A\) "].

Short Answer

Expert verified
If \(A\) attracts \(B\), then \(B\) attracts \(A\) by Bayes' theorem and inequality transformations.

Step by step solution

01

Express Conditional Probabilities with Bayes' Theorem

We start by using Bayes' Theorem, which states that \( P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)} \). Similarly, \( P(B \mid A) = \frac{P(A \mid B) \cdot P(B)}{P(A)} \).
02

Utilize the Given Inequality

We know from the problem that \( P(B \mid A) > P(B) \). This implies that \( \frac{P(B \mid A) \cdot P(A)}{P(B)} > P(A) \). This means that the presence of event A increases the probability of event B occurring.
03

Simplify Inequality with Bayes' Formula

Substitute \( P(B \mid A) > P(B) \) into Bayes' theorem expression, \( \frac{P(A \mid B) \cdot P(B)}{P(A)} = P(B \mid A) \). With the inequality \( P(B \mid A) > P(B) \), we have \( \frac{P(A \mid B) \cdot P(B)}{P(A)} > P(B) \cdot P(A) \).
04

Rearrange and Solve the Inequality

Rearrange the inequality from Step 3: \( \frac{P(A \mid B)}{P(A)} > 1 \). Multiply both sides by \( P(A) \), giving us \( P(A \mid B) > P(A) \).
05

Conclusion

From our rearranged inequality, we see that \( P(A \mid B) > P(A) \), proving that if \( A \) attracts \( B \), then \( B \) also attracts \( A \).

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Key Concepts

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

Bayes' Theorem
Bayes' Theorem is a foundational concept in probability theory that relates the conditional and marginal probabilities of random events. It offers a way to update our belief about the probability of an event based on new evidence.
If you know certain probabilities such as the likelihood of observing evidence given that a hypothesis is true, Bayes' Theorem allows you to compute the probability of the hypothesis given the evidence. This relationship is expressed using the formula:
  • \(P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}\)
This formula states that the probability of event \(A\) occurring given that event \(B\) has occurred is equal to the probability of \(B\) occurring given \(A\), times the probability of \(A\) occurring, all divided by the probability of \(B\) occurring.
This theorem is widely used across various fields, such as in decision making, diagnostics, and machine learning, whenever we need to incorporate new data or evidence to reassess probabilities.
Inequality of Probabilities
The Inequality of Probabilities refers to the comparison between conditional and unconditional probabilities. If we say \(P(B \mid A) > P(B)\), it means that the occurrence of event \(A\) increases the likelihood of event \(B\).
In other words, knowing that \(A\) has occurred provides additional information that makes \(B\) more likely than it would be on its own.
This idea is key to understanding dependent events, where the probability of an event is influenced by the occurrence of another event.
In mathematical terms, when we have \(P(B \mid A) > P(B)\), we realize that event \(A\) positively influences or conditions event \(B\), thereby increasing the probability of \(B\). This relationship highlights the importance of context and prior information in probability assessments.
Event Attraction in Probability
Event Attraction in Probability involves understanding how one event can affect the likelihood of another. When we say "\(A\) attracts \(B\)", it is based on the idea that the occurrence of \(A\) increases the probability of \(B\).
If \(P(B \mid A) > P(B)\), then \(A\) has a positive impact on \(B\), making \(B\) more likely once \(A\) occurs.
Conversely, if \(B\) attracts \(A\), then \(P(A \mid B) > P(A)\), indicating that the occurrence of \(B\) similarly boosts the chance of \(A\), demonstrating mutual attraction.
This dual attraction showcases a symmetrical relationship in probability where influence is reciprocal.
  • Example: Knowing that it will rain (event \(A\)) increases the likelihood of carrying an umbrella (event \(B\)). Similarly, if we see people carrying umbrellas (event \(B\)), we might infer it might rain (event \(A\)).
Understanding such interactions is crucial in fields like statistics, risk management, and decision sciences, where predicting outcomes based on interconnected events is essential.

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Most popular questions from this chapter

A boiler has five identical relief valves. The probability that any particular valve will open on demand is 95 . Assuming independent operation of the valves, calculate \(P\) (at least one valve opens) and \(P\) (at least one valve fails to open).

Allan and Beth currently have \(\$ 2\) and \(\$ 3\), respectively. A fair coin is tossed. If the result of the toss is \(\mathrm{H}\), Allan wins \(\$ 1\) from Beth, whereas if the coin toss results in \(\mathrm{T}\), then Beth wins \(\$ 1\) from Allan. This process is then repeated, with a coin toss followed by the exchange of \(\$ 1\), until one of the two players goes broke (one of the two gamblers is ruined). We wish to determine \(a_{2}=P\) (Allan is the winner \(\mid\) he starts with \(\$ 2\) ) To do so, let's also consider \(a_{i}=P\) (Allan wins | he starts with \(\$ i\) ) for \(i=0,1,3,4\), and 5 . a. What are the values of \(a_{0}\) and \(a_{5}\) ? b. Use the law of total probability to obtain an equation relating \(a_{2}\) to \(a_{1}\) and \(a_{3}\). [Hint: Condition on the result of the first coin toss, realizing that if it is a \(\mathrm{H}\), then from that point Allan starts with \$3.] c. Using the logic described in (b), develop a system of equations relating \(a_{i}(i=1,2,3,4)\) to \(a_{i-1}\) and \(a_{i+1}\). Then solve these equations. [Hint: Write each equation so that \(a_{i}-a_{i-1}\) is on the left hand side. Then use the result of the first equation to express each other \(a_{i}-a_{i-1}\) as a function of \(a_{1}\), and add together all four of these expressions \((i=2,3,4,5)\).] d. Generalize the result to the situation in which Allan's initial fortune is \(\$ a\) and Beth's is \(\$ b\). Note: The solution is a bit more complicated if \(p=P(\) Allan wins \(\$ 1) \neq .5 .\)

Suppose an individual is randomly selected from the population of all adult males living in the United States. Let \(A\) be the event that the selected individual is over \(6 \mathrm{ft}\) in height, and let \(B\) be the event that the selected individual is a professional basketball player. Which do you think is larger, \(P(A \mid B)\) or \(P(B \mid A)\) ? Why?

One box contains six red balls and four green balls, and a second box contains seven red balls and three green balls. A ball is randomly chosen from the first box and placed in the second box. Then a ball is randomly selected from the second box and placed in the first box. a. What is the probability that a red ball is selected from the first box and a red ball is selected from the second box? b. At the conclusion of the selection process, what is the probability that the numbers of red and green balls in the first box are identical to the numbers at the beginning?

A certain system can experience three different types of defects. Let \(A_{i}(i=1,2,3)\) denote the event that the system has a defect of type \(i\). Suppose that $$ \begin{aligned} &P\left(A_{1}\right)=.12 \quad P\left(A_{2}\right)=.07 \quad P\left(A_{3}\right)=.05 \\ &P\left(A_{1} \cup A_{2}\right)=.13 \quad P\left(A_{1} \cup A_{3}\right)=.14 \\\ &P\left(A_{2} \cup A_{3}\right)=.10 \quad P\left(A_{1} \cap A_{2} \cap A_{3}\right)=.01 \end{aligned} $$ a. What is the probability that the system does not have a type 1 defect? b. What is the probability that the system has both type 1 and type 2 defects? c. What is the probability that the system has both type 1 and type 2 defects but not a type 3 defect? d. What is the probability that the system has at most two of these defects?

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