The joint density function of the random variables X and Y is given to the right. (a) Show that X and Y are not independent. (b) Find P(X>0.2 | Y=0.6). (a) Select the correct choice below and fill in the answer box to complete your choice. O A. f(x,y) h(y) O B. O C. O D. Since f(xly)=- Since f(xly) = Since f(xly) = f(x,y) h(y) f(x,y) h(y) f(x,y) h(y) Since f(xly)=- f(x,y)= 12x², 00.2 | Y=0.6)=[ (Simplify your answer. Round to six decimal places as needed.) , for 0

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The joint density function of the random variables X and Y is given to the
right.
(a) Show that X and Y are not independent.
(b) Find P(X>0.2 | Y=0.6).
(a) Select the correct choice below and fill in the answer box to complete your choice.
O A.
O B.
O C.
O D.
Since f(xly) =
Since f(xly) =
f(x,y)
h(y)
Since f(xly) =
f(x,y)
h(y)
f(x,y)
Since f(xly) = -
h(y)
f(x,y)
h(y)
f(x,y) =
12x²,0<x<1, 0 <y <1-x
0₁
, for 0 <x< 1-y, is constant, X and Y are not independent.
(b) P(X>0.2 | Y=0.6)=
(Simplify your answer. Round to six decimal places as needed.)
elsewhere
, for 0 <y <1-x, involves the variable x, X and Y are not independent.
, for 0 <x< 1-y, involves the variable y, X and Y are not independent.
for 0 <x< 1-y, is a function of only the variable x, X and Y are not independent.
Transcribed Image Text:The joint density function of the random variables X and Y is given to the right. (a) Show that X and Y are not independent. (b) Find P(X>0.2 | Y=0.6). (a) Select the correct choice below and fill in the answer box to complete your choice. O A. O B. O C. O D. Since f(xly) = Since f(xly) = f(x,y) h(y) Since f(xly) = f(x,y) h(y) f(x,y) Since f(xly) = - h(y) f(x,y) h(y) f(x,y) = 12x²,0<x<1, 0 <y <1-x 0₁ , for 0 <x< 1-y, is constant, X and Y are not independent. (b) P(X>0.2 | Y=0.6)= (Simplify your answer. Round to six decimal places as needed.) elsewhere , for 0 <y <1-x, involves the variable x, X and Y are not independent. , for 0 <x< 1-y, involves the variable y, X and Y are not independent. for 0 <x< 1-y, is a function of only the variable x, X and Y are not independent.
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