/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Suppose you choose two numbers \... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose you choose two numbers \(x\) and \(y\), independently at random from the interval \([0,1] .\) Given that their sum lies in the interval [0,1] , find the probability that (a) \(|x-y|<1\). (b) \(x y<1 / 2\). (c) \(\max \\{x, y\\}<1 / 2\). (d) \(x^{2}+y^{2}<1 / 4\). (e) \(x>y\)

Short Answer

Expert verified
(a) 1; (b) 1; (c) 1/2; (d) \(\pi/8\); (e) 1/2.

Step by step solution

01

Understanding the Geometric Probability Setup

Since we are dealing with random choices of \(x\) and \(y\) from the interval \([0,1]\), we can think of this as choosing points uniformly at random from the square \([0,1] \times [0,1]\) on a coordinate plane. However, given that their sum \(x + y\) lies in the interval \([0, 1]\), we are limited to the triangular region that has vertices at points (0,0), (1,0), and (0,1). This is the region of interest for all parts.
02

Step 2(a): Calculating Probability for \(|x-y|

The condition \(|x-y| < 1\) covers the entire triangular region because \(x\) and \(y\) being in \([0,1]\) makes \(|x-y|\) always less than 1. Therefore, we are working with the entire area of the triangle. The area of the triangle is \(\frac{1}{2}\times 1 \times 1 = \frac{1}{2}\). Thus, the probability is \(\frac{\text{Area with } |x-y|<1}{\text{Area of the entire region}} = 1\).
03

Step 3(b): Calculating Probability for \(x y

For \(x y < \frac{1}{2}\), we check which parts of the triangle fall below the hyperbola \(xy = \frac{1}{2}\). In the region \(x+y \leq 1\), the constraint \(xy < \frac{1}{2}\) will be satisfied entirely because \(xy\) cannot be \(\frac{1}{2}\) due to the boundary \(x + y = 1\). Therefore, the probability is 1.
04

Step 4(c): Calculating Probability for \(\max{x,y}

The condition \(\max{x,y} < \frac{1}{2}\) means both \(x < \frac{1}{2}\) and \(y < \frac{1}{2}\). This means we are restricted to the square region bounded at the top right by \((\frac{1}{2}, \frac{1}{2})\), which has an area of \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\). Therefore, the probability is \(\frac{\text{Area of small square}}{\text{Area of triangle}} = \frac{1/4}{1/2} = \frac{1}{2}\).
05

Step 5(d): Calculating Probability for \(x^{2}+y^{2}

For \(x^{2}+y^{2}<\frac{1}{4}\), we are finding the area within a quarter circle of radius \(\frac{1}{2}\) (since \(x, y \leq 1\)) under the line \(x+y = 1\). The quarter circle fits entirely within the triangle and has an area of \(\frac{\pi \times (1/2)^2}{4} = \frac{\pi}{16}\). The probability is then \(\frac{\frac{\pi}{16}}{\frac{1}{2}} = \frac{\pi}{8}\).
06

Step 6(e): Calculating Probability for \(x>y\)

For \(x>y\), we find the region where the horizontal component \(x\) exceeds the vertical component \(y\). Given the uniform distribution over the region \(x + y \leq 1\), the line \(x = y\) divides the region into two equal areas. Thus, the probability is \(\frac{1}{2}\).

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Key Concepts

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

Random Variables
Random variables are essential in probability theory because they allow us to model and calculate unpredictable outcomes. They are used to represent numerical outcomes from random experiments or processes. In this context, choosing numbers \(x\) and \(y\) from the interval \([0,1]\) creates two random variables, typically denoted \(X\) and \(Y\). These variables are independent, meaning the selection of \(x\) does not affect the selection of \(y\) and vice versa. This independence is crucial because it simplifies how we analyze the probability of combined conditions, such as sums or maximum values, as seen in the exercise.
\(X\) and \(Y\) are uniform random variables, meaning each number in \([0,1]\) is equally likely to be chosen. This uniformity ensures an unbiased representation across the interval and simplifies the geometric probability assessments discussed in the problem.
Probability Calculation
Probability calculation is about finding the likelihood of an event occurring. In geometric probability, we calculate this by comparing areas. Each condition in the exercise can be seen as a region within the triangular space defined by \(x+y \leq 1\). For instance:
  • In part (a), the condition \(|x-y|<1\) is always true within the triangle. Therefore, the probability is 1 since the entire region satisfies \(|x-y|<1\).
  • For condition \(xy<\frac{1}{2}\) in part (b), the entire triangular region naturally falls below the curve \(xy=\frac{1}{2}\) because of the boundary \(x+y=1\), resulting again in a probability of 1.
  • In part (c), the constraint \(\max{x,y}<\frac{1}{2}\) limits the solution to a square in the bottom-left of the region. Calculating the area ratio gives us a probability of \(\frac{1}{2}\).
Calculating these probabilities is done by measuring how much of the originally defined space actually satisfies the conditions. These geometry-based probabilities offer a visual and intuitive way to understand the likelihood of complex events.
Triangular Region
The triangular region arises naturally from the condition \(x+y\leq 1\). This condition restricts the choices of \(x\) and \(y\) to a triangle with vertices at (0,0), (1,0), and (0,1). Hence, the area of the triangle plays a significant role in determining the probabilities of each condition.
In geometric probability, determining areas like this triangle guides our understanding of how likely different outcomes are in the context of random variables. The triangle limits the area we are interested in, setting the stage for other calculations. For example, understanding that this triangle is our base region (with an area of \(\frac{1}{2}\)) is crucial for normalizing probabilities when we calculate the probability of events within this boundary.
Coordinate Plane Analysis
Analyzing the coordinate plane involves assessing the relationship between \(x\) and \(y\), particularly when they are plotted on a 2D graph. With \(x\) and \(y\) both ranging from 0 to 1, we initially start with a unit square. However, the condition \(x+y \leq 1\) confines all points to the triangular region as discussed.
Coordinate plane analysis lets us visualize and analyze regions like:
  • The effectively half region defined by \(x>y\) where each half of the triangle split by the line \(x=y\) represents equal probabilities, suggesting symmetry and leading to a probability of \(\frac{1}{2}\) in step 6(e).
  • Areas below certain curves or hyperbolas, such as \(xy=\frac{1}{2}\), demonstrate how random variables interact based on different constraints.
This kind of analysis is powerful as it transitions abstract conditions into tangible spaces on the coordinate plane, enabling clearer understanding and precise probability calculations.

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Most popular questions from this chapter

A coin is in one of \(n\) boxes. The probability that it is in the \(i\) th box is \(p_{i}\). If you search in the \(i\) th box and it is there, you find it with probability \(a_{i}\). Show that the probability \(p\) that the coin is in the \(j\) th box, given that you have looked in the \(i\) th box and not found it, is $$ p=\left\\{\begin{array}{cc} p_{j} /\left(1-a_{i} p_{i}\right), & \text { if } j \neq i \\ \left(1-a_{i}\right) p_{i} /\left(1-a_{i} p_{i}\right), & \text { if } j=i \end{array}\right. $$

In the problem of points, discussed in the historical remarks in Section 3.2, two players, A and B, play a series of points in a game with player A winning each point with probability \(p\) and player \(\mathrm{B}\) winning each point with probability \(q=1-p\). The first player to win \(N\) points wins the game. Assume that \(N=3 .\) Let \(X\) be a random variable that has the value 1 if player A wins the series and 0 otherwise. Let \(Y\) be a random variable with value the number of points played in a game. Find the distribution of \(X\) and \(Y\) when \(p=1 / 2\). Are \(X\) and \(Y\) independent in this case? Answer the same questions for the case \(p=2 / 3\).

Suppose you toss a dart at a circular target of radius 10 inches. Given that the dart lands in the upper half of the target, find the probability that (a) it lands in the right half of the target. (b) its distance from the center is less than 5 inches. (c) its distance from the center is greater than 5 inches. (d) it lands within 5 inches of the point (0,5) .

You are given two urns each containing two biased coins. The coins in urn I come up heads with probability \(p_{1}\), and the coins in urn II come up heads with probability \(p_{2} \neq p_{1}\). You are given a choice of (a) choosing an urn at random and tossing the two coins in this urn or (b) choosing one coin from each urn and tossing these two coins. You win a prize if both coins turn up heads. Show that you are better off selecting choice (a).

Probability theory was used in a famous court case: People v. Collins. \(^{10}\) In this case a purse was snatched from an elderly person in a Los Angeles suburb. A couple seen running from the scene were described as a black man with a beard and a mustache and a blond girl with hair in a ponytail. Witnesses said they drove off in a partly yellow car. Malcolm and Janet Collins were arrested. He was black and though clean shaven when arrested had evidence of recently having had a beard and a mustache. She was blond and usually wore her hair in a ponytail. They drove a partly yellow Lincoln. The prosecution called a professor of mathematics as a witness who suggested that a conservative set of probabilities for the characteristics noted by the witnesses would be as shown in Table 4.5 . The prosecution then argued that the probability that all of these characteristics are met by a randomly chosen couple is the product of the probabilities or \(1 / 12,000,000\), which is very small. He claimed this was proof beyond a reasonable doubt that the defendants were guilty. The jury agreed and handed down a verdict of guilty of second-degree robbery. $$ \begin{aligned} &\begin{array}{lc} \text { man with mustache } & 1 / 4 \\ \text { girl with blond hair } & 1 / 3 \\ \text { girl with ponytail } & 1 / 10 \\ \text { black man with beard } & 1 / 10 \\ \text { interracial couple in a car } & 1 / 1000 \\ \text { partly yellow car } & 1 / 10 \end{array}\\\ &\text { Table 4.5: Collins case probabilities. } \end{aligned} $$ If you were the lawyer for the Collins couple how would you have countered the above argument? (The appeal of this case is discussed in Exercise 5.1.34.)

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