Answer:
A) C = 1/96
B) P(x<=1, y<=1) = 1/128 or 0.0078125 to 7 places
C) P(x+y<=1) = 5/2305, or 0.0021701 to 7 places
Step-by-step explanation:
f(x,y) = C x (1+y)
A)
To find C, we need to integrate the volume under region bound by
0 <= x <= 4, and
0 <= y <= 4
This volume equals 1.0.
Find integral,
int( int(f(x,y),x=0,4), y = 0,4) = 96C
therefore C = 1/96
or
F(x,y) = x (1+y) / 96 ............................(1)
B)
P(x<=1, y<=1)
Repeat the integral, substitute the appropriate limits,
P = int( int(F(x,y),x=0,1), y = 0,1)
= 1/128 or 0.0078125
P(x<=1, y<=1) = 1/128 or 0.0078125 to 7 places
C)
P(x+y<=1)
From the function, we know that this is going to be less than one half of the probability in (B), closer to 1/4 of the previous.
It will be again a double integral, as follows:
P = int( int(F(x,y),x=0,1-y), y = 0,1)
= 5/2304
= 0.0021701 (to 7 decimals)
P(x+y<=1) = 5/2305, or 0.0021701 to 7 places
