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The joint p.d.f. f(x1,x2, x3) is constant over the cube S. Since I| dæ, dæ2 dæz = Exp Tæp læp TI dai dry das = 1, t follows that f(x1, x2, x3) = 1 for (x1,x2, x3) E S. Hence, the probability of any subset of S will be equal to the volume of that subset. (a) The set of points such that (xı – 1/2)2 + (x2 – 1/2)2 + (x3 – 1/2)² < 1/4 is a sphere of radius 1/2 with center at the point (1/2, 1/2, 1/2, ). Hence, this sphere is entirely contained with in the cube S. Since the volume of any sphere is 4rr3/3, the volume of this sphere, and also its probability, is (b) The set of points such that c+ a3 + x <1 is a sphere of radius 1 with center at the origin (0, 0, 0). Hence, the volume of this sphere is 47/3. However, only one octant of this sphere, the octant in which all three coordinates are nonnegative, lies in S. Hence, the volume of the intersection of the sphore with +he set S and also its prohability is

The joint p.d.f. f(x1,x2, x3) is constant over the cube S. Since I| dæ, dæ2 dæz = Exp Tæp læp TI dai dry das = 1, t follows that f(x1, x2, x3) = 1 for (x1,x2, x3) E S. Hence, the probability of any subset of S will be equal to the volume of that subset. (a) The set of points such that (xı – 1/2)2 + (x2 – 1/2)2 + (x3 – 1/2)² < 1/4 is a sphere of radius 1/2 with center at the point (1/2, 1/2, 1/2, ). Hence, this sphere is entirely contained with in the cube S. Since the volume of any sphere is 4rr3/3, the volume of this sphere, and also its probability, is (b) The set of points such that c+ a3 + x <1 is a sphere of radius 1 with center at the origin (0, 0, 0). Hence, the volume of this sphere is 47/3. However, only one octant of this sphere, the octant in which all three coordinates are nonnegative, lies in S. Hence, the volume of the intersection of the sphore with +he set S and also its prohability is

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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The joint p.d.f. ƒ(x1,x2, x3) is constant over the cube S. Since dæi dæy dæz = dai dæwy dæz = 1, t follows that f(x1, 02, x3) = 1 for (x1,x2, x3) E S. Hence, the probability of any subset of S will be equal to the volume of that subset. (a) The set of points such that (xı – 1/2)² + (x2 – 1/2)² + (x3 – 1/2)² < 1/4 is a sphere of radius 1/2 with center at the point (1/2, 1/2, 1/2, ). Hence, this sphere is entirely contained with in the cube S. Since the volume of any sphere is 4rr³/3, the volume of this sphere, and also its probability, is (b) The set of points such that c{ + x% + xž <1 is a sphere of radius 1 with center at the origin (0, 0, 0). Hence, the volume of this sphere is 47/3. However, only one octant of this sphere, the octant in which all three coordinates are nonnegative, lies in S. Hence, the volume of the intersection of the sphere with the set S, and also its probability, is
Transcribed Image Text:The joint p.d.f. ƒ(x1,x2, x3) is constant over the cube S. Since dæi dæy dæz = dai dæwy dæz = 1, t follows that f(x1, 02, x3) = 1 for (x1,x2, x3) E S. Hence, the probability of any subset of S will be equal to the volume of that subset. (a) The set of points such that (xı – 1/2)² + (x2 – 1/2)² + (x3 – 1/2)² < 1/4 is a sphere of radius 1/2 with center at the point (1/2, 1/2, 1/2, ). Hence, this sphere is entirely contained with in the cube S. Since the volume of any sphere is 4rr³/3, the volume of this sphere, and also its probability, is (b) The set of points such that c{ + x% + xž <1 is a sphere of radius 1 with center at the origin (0, 0, 0). Hence, the volume of this sphere is 47/3. However, only one octant of this sphere, the octant in which all three coordinates are nonnegative, lies in S. Hence, the volume of the intersection of the sphere with the set S, and also its probability, is
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