/*! 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 4 Find \(P\left(0... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 4

Find \(P\left(0

Short Answer

Expert verified
\(\frac{4}{81}\)

Step by step solution

01

Set up the Double Integral

We want to find \(P\left(0<X_{1}<\frac{1}{3}, 0<X_{2}<\frac{1}{3}\right)\). This can be calculated by integrating the joint pdf over the specified range. Thus, set up the double integral as follows: \[\int_{0}^{1/3}\int_{0}^{1/3}4x_{1}(1-x_{2})dx_{1}dx_{2}\].
02

Integrate with Respect to \(x_{1}\)

To start with, integrate the inner integral with respect to \(x_{1}\). \[\int_{0}^{1/3}4x_{1}(1-x_{2})dx_{1} = 4(1-x_{2})(\frac{1}{2}x_{1}^2)\Bigg|_{0}^{1/3} = \frac{4}{18}(1-x_{2}) = \frac{2}{9}(1-x_{2}).\] This gives us the integral \( \frac{2}{9}\int_{0}^{1/3}(1-x_{2})dx_{2}\).
03

Integrate with Respect to \(x_{2}\)

Now, it's time to integrate the result with respect to \(x_{2}\). This gives us: \[\frac{2}{9} \int_{0}^{1/3}(1-x_{2})dx_{2} = \frac{2}{9}[(x_{2}-\frac{1}{2}x_{2}^{2})]_{0}^{1/3} = \frac{2}{9}[(1/3 - 1/18)] = \frac{2}{9}*(4/18)=\frac{4}{81}.\]
04

Interpret the Result

The obtained result, \(\frac{4}{81}\), represents the probability of the given event. This value lies between 0 and 1 which validates the probability.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}\) have the joint pdf \(f\left(x_{1}, x_{2}\right)=1 / \pi, 0

Let \(X\) and \(Y\) have the joint pdf \(f(x, y)=3 x, 0

Let \(A_{1}=\\{(x, y): x \leq 2, y \leq 4\\}, A_{2}=\\{(x, y): x \leq 2, y \leq 1\\}, A_{3}=\) \(\\{(x, y): x \leq 0, y \leq 4\\}\), and \(A_{4}=\\{(x, y): x \leq 0 y \leq 1\\}\) be subsets of the space \(\mathcal{A}\) of two random variables \(X\) and \(Y\), which is the entire two-dimensional plane. If \(P\left(A_{1}\right)=\frac{7}{8}, P\left(A_{2}\right)=\frac{4}{8}, P\left(A_{3}\right)=\frac{3}{8}\), and \(P\left(A_{4}\right)=\frac{2}{8}\), find \(P\left(A_{5}\right)\), where \(A_{5}=\\{(x, y): 0

Let \(X\) and \(Y\) be random variables with the space consisting of the four points \((0,0),(1,1),(1,0),(1,-1)\). Assign positive probabilities to these four points so that the correlation coefficient is equal to zero. Are \(X\) and \(Y\) independent?

If the correlation coefficient \(\rho\) of \(X\) and \(Y\) exists, show that \(-1 \leq \rho \leq 1\). Hint: Consider the discriminant of the nonnegative quadratic function $$ h(v)=E\left\\{\left[\left(X-\mu_{1}\right)+v\left(Y-\mu_{2}\right)\right]^{2}\right\\} $$ where \(v\) is real and is not a function of \(X\) nor of \(Y\).
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