   Chapter 15, Problem 51RE

Chapter
Section
Textbook Problem

The joint density function for random variables X and Y is f ( x ,   y )   = { C ( x + y ) if 0 ≤ x ≤ 3 , 0 ≤ y ≤ 2 0 otherwise (a) Find the value of the constant C.(b) Find P(X ≤ 2, Y ≥ 1).(c) Find P(X + Y ≤ 1).

(a)

To determine

To find: The value of C.

Explanation

Given:

The joint density function, f(x,y)={C(x+y) , if 0x3,0y2       0       ,  otherwise.

The region D is  0x3,0y2.

Property used:

The double integral value of the given function over D is always 1 that is f(x,y)dA=1.

Formula used:

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (1)

Calculation:

By the property mentioned above, the value of f(x,y)dA becomes,

f(x,y)dA=00f(x,y)dydx+0302f(x,y)dydx+32f(x,y)dydx=1

From the given condition, the value of f(x,y)dA is,

f(x,y)dA=0+0302f(x,y)dydx+0=0302f(x,y)dydx=1

Obtain the value of the double integral

(b)

To determine

To find: The value of P(X2,Y1).

(c)

To determine

To find: The value of P(X+Y1).

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