document.write( "Question 1105749: given the following joint probability distribution:
\n" ); document.write( "f(x,y)=x+y/30 , x=0,1,2,3;y=0,1,2
\n" ); document.write( "1)Find P(X+Y=4) and P(X>Y).
\n" ); document.write( "2)Check whether X and Y are independent random variables. \n" ); document.write( "
Algebra.Com's Answer #720651 by Boreal(15235)\"\" \"About 
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I'm assuming f (x,y)=(x+y)/30
\n" ); document.write( "probability= 0 at (0, 0)
\n" ); document.write( "probability=1/30 at (1, 0) and (0, 1)
\n" ); document.write( "at (2, 0), (1, 1), (0, 2) probability is 2/30
\n" ); document.write( "at (3, 0), (2, 1), (1, 2) probability is 3/30
\n" ); document.write( "at (3, 1) and (2, 2), probability is 4/30
\n" ); document.write( "at (3, 2), probability is 5/30
\n" ); document.write( "Those probabilities do add to 1
\n" ); document.write( "x+y=4 occurs at allowable (2, 2) and (3, 1)
\n" ); document.write( "That has probability of 8/30
\n" ); document.write( "x>y occurs at (1, 0), (2, 0), (3, 0), (2, 1), (3, 1), and (3, 2)
\n" ); document.write( "That probability is 18/30 or 3/5
\n" ); document.write( "If independent, probability (x and y)=p(x)*p (y)
\n" ); document.write( "The probability of (2, 2), which is 4/30 or 0.133=probability x=2, which is 11/30* probability y=2, which is 14/30.
\n" ); document.write( "That product is 154/900 or 0.171.
\n" ); document.write( "0.133 is not equal to 0.171, so not independent. \n" ); document.write( "
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