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
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
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