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Chapter 2: Problem 31

A document is equally likely to be in any of three boxfiles. A search of the \(i\) th box will discover the document (if it is indeed there) with probability \(p_{i}\). What is the probability that the document is in the first box: (a) Given that I have searched the first box once and not found it? (b) Given that I have searched the first box twice and not found it? (c) Given that I have searched all three boxes once and not found it? Assume searches are independent.

Short Answer

Expert verified
(a) \( \frac{1-p_1}{3-p_1}\); (b) \( \frac{(1-p_1)^2}{3-2p_1+(1-p_1)^2}\); (c) \( \frac{1-p_1}{3-p_1-p_2-p_3} \).

Step by step solution

01

Understand the Problem

We have three boxfiles and a document that can be in any of these with an equal probability of \( \frac{1}{3} \). We also have a probability \( p_i \) that searching the \( i^{th} \) box will find the document if it is there.
02

Identify the Basic Probability Elements

Define events: \( A_i \) as the document being in the \( i^{th} \) box. We also define \( S_i \) as the event that a document search in the \( i^{th} \) box comes up empty-handed. We want to find \( P(A_1 \mid S_1) \), \( P(A_1 \mid S_1^2) \), and \( P(A_1 \mid S_1 \cap S_2 \cap S_3) \).
03

Use Bayes' Theorem for Part (a)

For part (a), we apply Bayes' Theorem: \( P(A_1 \mid S_1) = \frac{P(S_1 \mid A_1)P(A_1)}{P(S_1)} \). Since \( P(S_1 \mid A_1) = 1 - p_1 \) and \( P(S_1) = \frac{1-p_1}{3} + \frac{1}{3}(1) + \frac{1}{3}(1) \), we substitute these into the formula: \( P(A_1 \mid S_1) = \frac{\frac{1-p_1}{3}}{\frac{3-p_1}{3}} = \frac{1-p_1}{3-p_1} \).
04

Repeat Bayes' Theorem for Part (b)

For part (b), when the first box is searched twice, the probability becomes \( P(S_1^2 \mid A_1) = (1-p_1)^2 \). Bayes' theorem gives us: \( P(A_1 \mid S_1^2) = \frac{\frac{(1-p_1)^2}{3}}{\frac{(1-p_1)^2}{3} + 1 - \frac{1}{3}(1-p_1)^2} = \frac{(1-p_1)^2}{3-2p_1 + (1-p_1)^2} \).
05

Compute Probability for Part (c)

For part (c), search each box once without finding the document: \( P(S_1 \cap S_2 \cap S_3 \mid A_1) = (1-p_1) \) and \( P(S_1 \cap S_2 \cap S_3) = \frac{1-p_1}{3} + \frac{1-p_2}{3} + \frac{1-p_3}{3} \). Thus, \( P(A_1 \mid S_1 \cap S_2 \cap S_3) = \frac{\frac{1-p_1}{3}}{\frac{1-p_1 + 1-p_2 + 1-p_3}{3}} = \frac{1-p_1}{3-p_1-p_2-p_3} \).

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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 fundamental concept in probability that helps us update the probability of an event based on new evidence. It's especially useful in situations where we need to make predictions and decisions under uncertainty. Bayes' theorem relates the conditional and marginal probabilities of random events. It is expressed as:
\[P(A \,|\, B) = \frac{P(B \,|\, A) \cdot P(A)}{P(B)}\]Here, \(P(A \,|\, B)\) represents the probability of event \(A\) occurring given that \(B\) is true. This allows us to update our belief about \(A\) based on the evidence \(B\). In our problem, this theorem is used to determine the probability of the document being in a particular box after a search has been conducted. When we search and don't find the document, Bayes' theorem helps us revise the probability that the document was there all along. By repeating this process for each instance, we can adjust our expectations and make better decisions based on new data or searches. In cases where probabilities are independent, Bayes' theorem is sometimes simplified to ease calculations and interpretations.
independent events
In probability theory, events are independent when the occurrence of one does not affect the probability of another occurring. For example, if you roll a die, getting a 3 doesn't change your chances of rolling a 5 next time. This concept is crucial in our scenario. The searches of the boxes are independent events, meaning the result of searching one box doesn't alter the result of searching another. Here's why it's important:
  • Independence simplifies probability calculations. We multiply the probabilities of individual events to get the combined probability.
  • It allows us to use straightforward probabilities in Bayes' theorem.
  • Assumptions about independence are necessary to compute the conditional probability accurately.
By considering the searches as independent, we're saying that not finding the document in a search doesn't improve or worsen the chances on the next box search. This independence is factored into the calculations in parts (a), (b), and (c) of the original problem, ensuring the adjustments in probabilities are correctly applied and making sure calculations stay manageable even as sequences of searches increase.
probability calculation
Probability calculations form the backbone of solving any statistical problem. They help us determine the likelihood of various outcomes and make smarter decisions. Fundamental manipulations and properties of probabilities include:
  • The sum of probabilities for all possible outcomes equals 1. If you are sure of something happening, the probability is 1.
  • While calculating conditional probabilities, remember: \(P(A \,|\,B) eq P(B \,|\,A)\).
  • Using conditional probabilities involves breaking down complex problems into more manageable steps, as seen with Bayes' theorem.
In our exercise, calculating the probability that the document is in the first box after failed searches involves:
1. Calculating initial probabilities that the document resides in each of the three boxes (equal probabilities of \(\frac{1}{3}\)).
2. Adjusting these probabilities based on the new information (e.g., failed searches), using Bayes' theorem.
3. Incorporating independence of searches (they don't affect each other's outcomes) helps keep things straightforward.
These principles apply naturally to both the given exercise and many real-world scenarios, helping make informed decisions based on probabilistic reasoning.

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Most popular questions from this chapter

Suppose that any child is male with probability \(p\) or female with probability \(1-p\), independently of other children. In a family with four children, let \(A\) be the event that there is at most one girl, and \(B\) the event that there are children of both sexes. Show that there is a value of \(p\), with \(0

Simpson's Paradox Two drugs are being tested. Of 200 patients given drug \(A, 60\) are cured; and of 1100 given drug \(B, 170\) are cured. If we assume a homogeneous group of patients, find the probabilities of successful treatment with \(A\) or \(B\). Now closer investigation reveals that the 200 patients given drug \(A\) were in fact 100 men, of whom 50 were cured, and 100 women of whom 10 were cured. Further, of the 1100 given drug \(B, 100\) were men of whom 60 were cured, and 1000 were women of whom 110 were cured. Calculate the probability of cure for men and women receiving each drug; note that \(B\) now seems better than \(A\). (Results of this kind indicate how much care is needed in the design of experiments. Note that the paradox was described by Yule in 1903 , and is also called the Yule-Simpson paradox.)

Box \(A\) contains three red balls and two white balls; box \(B\) contains two red balls and two white balls. A fair die is thrown. If the upper face of the die shows 1 or 2 , a ball is drawn at random from box \(A\) and put in box \(B\) and then a ball is drawn at random from box \(B\). If the upper face of the die shows \(3,4,5\) or 6 , a ball is drawn at random from box \(B\) and put in box \(A\), and then a ball is drawn at random from box \(A\). What are the probabilities (a) That the second ball drawn is white? (b) That both balls drawn are red? (c) That the upper face of the red die showed 3, given that one ball drawn is white and the other red?

Explain the following "paradox" posed by Lewis Carroll. We are provided with a supply of balls that are independently equally likely to be black or white. Proposition If an urn contains two such balls, then one is black and the other white. Proof Initially, \(\mathbf{P}(B B)=\mathbf{P}(B W)=\mathbf{P}(W B)=\mathbf{P}(W W)=\frac{1}{4}\). Add a black ball, so that now $$ \mathbf{P}(B B B)=\mathbf{P}(B B W)=\mathbf{P}(B W B)=\mathbf{P}(B W W)=\frac{1}{4} $$ Now pick a ball at random. By (2.1.3), $$ \mathbf{P}(\text { black ball drawn })=1 \cdot \frac{1}{4}+\frac{2}{3} \cdot \frac{1}{4}+\frac{2}{3} \cdot \frac{1}{4}+\frac{1}{3} \cdot \frac{1}{4}=\frac{2}{3} \text {. } $$ But if I pick a ball at random from three, with probability \(\frac{2}{3}\) of drawing a black ball, then two are black and one is white. Hence, before adding the black ball the urn contained one white ball and one black ball.

Suppose that parents are equally likely to have (in total) one, two, or three offspring. A girl is selected at random; what is the probability that the family includes no older girl? (Assume that children are independent and equally likely to be male or female.)
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