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ssrs 2008 r2 barcode font Random Variables and Probability Distributions in Software
CHAPTER 2 Random Variables and Probability Distributions Recognizing QR Code ISO/IEC18004 In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. QR Code ISO/IEC18004 Generator In None Using Barcode creator for Software Control to generate, create QRCode image in Software applications. Applications to geometric probability 2.31. A person playing darts finds that the probability of the dart striking between r and r Decode QR Code ISO/IEC18004 In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. Printing QRCode In C#.NET Using Barcode creation for VS .NET Control to generate, create Denso QR Bar Code image in Visual Studio .NET applications. P(r R r dr) c B1 r 2 a R dr
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Making QR Code 2d Barcode In VB.NET Using Barcode creation for .NET Control to generate, create QRCode image in Visual Studio .NET applications. Create GS1 128 In None Using Barcode generator for Software Control to generate, create UCC128 image in Software applications. Here, R is the distance of the hit from the center of the target, c is a constant, and a is the radius of the target (see Fig. 218). Find the probability of hitting the bull seye, which is assumed to have radius b. Assume that the target is always hit. Generating GTIN  12 In None Using Barcode generator for Software Control to generate, create UPCA Supplement 5 image in Software applications. Create Code128 In None Using Barcode creator for Software Control to generate, create Code 128B image in Software applications. The density function is given by f (r) Since the target is always hit, we have c B1 r 2 a R 1
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30 f (r) dr
2.32. Two points are selected at random in the interval 0 squares is less than 1.
3 b B1 2a 30
r 2 a R dr
b (3a2 b2) 2a3
1. Determine the probability that the sum of their
Let X and Y denote the random variables associated with the given points. Since equal intervals are assumed to have equal probabilities, the density functions of X and Y are given, respectively, by (1) f1(x) e 1 0 0 x 1 otherwise e f2 ( y) e 1 0 0 y 1 otherwise Then since X and Y are independent, the joint density function is given by (2) f (x, y) f1(x) f2(y) 1 0 0 x 1, 0 otherwise y 1 It follows that the required probability is given by (3) P(X2 Y2 1) 6 dx dy
where r is the region defined by x2 y2 1, x 0, y 0, which is a quarter of a circle of radius 1 (Fig. 219). Now since (3) represents the area of r, we see that the required probability is p > 4. CHAPTER 2 Random Variables and Probability Distributions
Fig. 219 Miscellaneous problems 2.33. Suppose that the random variables X and Y have a joint density function given by f (x, y) e c (2x 0 y) 2 x 6, 0 otherwise y 5
Find (a) the constant c, (b) the marginal distribution functions for X and Y, (c) the marginal density functions for X and Y, (d) P(3 X 4, Y 2), (e) P(X 3), (f) P(X Y 4), (g) the joint distribution function, (h) whether X and Y are independent. (a) The total probability is given by
6 5 2 y 6 c(2x
y) dx dy
3x 53 c 2xy
c 10x
For this to equal 1, we must have c
1 > 210. 25 dx 2
y2 5 2 dx 2 0
210c (b) The marginal distribution function for X is
F1(x) 3 ` v
f (u, v) du dv
x u 6
0 du dv
x 2x 2
2 5x 84 x 6 18 2 x 6
3 2 v 2u v du dv 0 210 2u v du dv 210
3 2 v The marginal distribution function for Y is
F2( y) 3 ` v
f (u, v) du dv
6 u 6 3 ` v
0 du dv
0 y2 0 0 5 y 5 3 0 v 2u v du dv 0 210 2u v du dv 210
16y 105 y
3 2 v CHAPTER 2 Random Variables and Probability Distributions
(c) The marginal density function for X is, from part (b), f1(x) d F (x) dx 1 e (4x 0 5)>84 2 x 6 otherwise The marginal density function for Y is, from part (b), f2( y) d F (y) dy 2 e (2y 0 1 4 210 3x
16)>105
0 y 5 otherwise y) dx dy 23 28 3 20
(d) (e) (f ) X P(X
4, Y 3) P(X
3 (2x 3 y 2
1 6 210 3x Y 4) 3 (2x 3 y 0
y) dx dy
6 f (x, y) dx dy
where r is the shaded region of Fig. 220. Although this can be found, it is easier to use the fact that P(X Y 4) 1 P(X Y 4) 1 6 f (x, y) dx dy where rr is the crosshatched region of Fig. 220. We have P(X Thus P(X Y 4) 33 > 35. Y 4) 1 4 210 3x 3 2 y y) dx dy
2 35 Fig. 220 Fig. 221 (g) The joint distribution function is
F(x, y) x, Y
3 ` v
f (u, v) du dv
In the uv plane (Fig. 221) the region of integration is the intersection of the quarter plane u x, v y and the rectangle 2 u 6, 0 v 5 [over which f (u, v) is nonzero]. For (x, y) located as in the figure, we have F(x, y) 3 2 v 2u v du dv 0 210
16y y 2 105
CHAPTER 2 Random Variables and Probability Distributions
When (x, y) lies inside the rectangle, we obtain another expression, etc. The complete results are shown in Fig. 222. (h) The random variables are dependent since f (x, y) 2 f1(x) f2 ( y) or equivalently, F(x, y) 2 F1(x)F2(y). 2.34. Let X have the density function f (x) Find a function Y e 6x (1 0 x) 0 x 1 otherwise
h(X) which has the density function g(y) e 12y 3(1 0 y 2) 0 y 1 otherwise
Fig. 222 We assume that the unknown function h is such that the intervals X x and Y y h(x) correspond in a oneone, continuous fashion. Then P(X x) P(Y y), i.e., the distribution functions of X and Y must be equal. Thus, for 0 x, y 1, 306u(1 or By inspection, x Y !X. y2 or y h(x) 3x2
u) du 2x3
3 3012v (1 v2) dv
!x is a solution, and this solution has the desired properties. Thus
2.35. Find the density function of U
Method 1 Let U XY and V
XY if the joint density function of X and Y is f(x, y). xy, v x or x v, y u > v. Then the Jacobian is given by
X, corresponding to which u 'x 'u 4 'y 'u 'x 'v 4 'y 'v
0 v 1 1 2 uv 2
CHAPTER 2 Random Variables and Probability Distributions
Thus the joint density function of U and V is g(u, v) from which the marginal density function of U is obtained as g(u) 3 `g(u, v) dv
1 u f v, v u vu
` u vu
Method 2 The distribution function of U is
G(u) For u B3
u f v, v dv
6 f (x, y) dx dy
xy u
0, the region of integration is shown shaded in Fig. 223. We see that
0 ` ` u>x
G(u) f (x, y) dyR dx
30 B 3 `f (x, y) dyR dx
Fig. 223 Fig. 224 Differentiating with respect to u, we obtain
g(u) The same result is obtained for u Fig. 224. 1 u x f x, x dx `
0, when the region of integration is bounded by the dashed hyperbola in
` 1 u 30 x f x, x dx
1 u f x, x dx u xu
2.36. A floor has parallel lines on it at equal distances l from each other. A needle of length a l is dropped at random onto the floor. Find the probability that the needle will intersect a line. (This problem is known as Buffon s needle problem.) Let X be a random variable that gives the distance of the midpoint of the needle to the nearest line (Fig. 224). Let be a random variable that gives the acute angle between the needle (or its extension) and the line. We denote by x and u any particular values of X and . It is seen that X can take on any value between 0 and l > 2, so that 0 x l > 2. Also can take on any value between 0 and p > 2. It follows that P(x i.e., the density functions of X and X x dx) 2 dx l P(u 2 > l, f2(u) 2 p du 1 du) 2 p du are given by f1(x)

