Question 1105749: given the following joint probability distribution:
f(x,y)=x+y/30 , x=0,1,2,3;y=0,1,2
1)Find P(X+Y=4) and P(X>Y).
2)Check whether X and Y are independent random variables.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! I'm assuming f (x,y)=(x+y)/30
probability= 0 at (0, 0)
probability=1/30 at (1, 0) and (0, 1)
at (2, 0), (1, 1), (0, 2) probability is 2/30
at (3, 0), (2, 1), (1, 2) probability is 3/30
at (3, 1) and (2, 2), probability is 4/30
at (3, 2), probability is 5/30
Those probabilities do add to 1
x+y=4 occurs at allowable (2, 2) and (3, 1)
That has probability of 8/30
x>y occurs at (1, 0), (2, 0), (3, 0), (2, 1), (3, 1), and (3, 2)
That probability is 18/30 or 3/5
If independent, probability (x and y)=p(x)*p (y)
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.
That product is 154/900 or 0.171.
0.133 is not equal to 0.171, so not independent.
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