Three balls are drawn at random with out replacement from a box containing 2 white, 3 red, 4 black balls. If x denotes the number of white balls drawn and y denotes the number of red balls drawn, find the joint probability distribution of (x,y).Also verify x and y are independent or not.
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my answer is yes they independen, in distribution ( 2,3)
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Given:
Total number of balls drawn = 3
Balls are drawn at random with out replacement from a box.
Number of White = 2
Number of Red = 3
Number of Black = 4
x = Number of white balls drawn
y = Number of red balls drawn.
To Find:
The joint probability distribution of (x,y).
Also verify x and y are independent or not.
Solution:
Joint Probability denotes the probability of the occurrence of both the events together.
Total number of balls = 9.
Total number of balls to be chosen = 3.
- Joint probability of x and y = P(X =x ,Y = y)
- X = { 0 , 1 , 2 }
- Y = { 0 , 1 , 2 , 3 }
P(X = x , Y = y ) = P(0,0) , P(0,1) , P(0,2), P(0,3) , P(1,0) , P(1,1) , P(1,2), P(2,0) ,P(2,1) .
- P(0,0) = All Black = 4/9 x 3/8 x 2/7 = 24 / 504
- P(0,1) = 2 B , 1 R = 4/9 x 3/8 x 3/7 = 36 / 504
- P(0,2) = 1B, 2R = 4/9 x 3/8 x 2/7 = 24 / 504
- P(0,3) = 3R = 3/9 x 2/8 x 1/7 = 6 / 504
- P(1,0) = 1White , 2 Black = 2/9 x 4/8 x 3/7 = 24 / 504
- P(1,1) = 1W,1R,1B = 2/9 x 3/8 x 4/7 = 24 / 504
- P(1,2) = 1W, 2R = 2/9 x 3/8 x 2/7 = 12/ 504
- P(2,0) = 2 white, 1 Black = 2/9 x 1/8 x 4/7 = 8 / 504
- P(2,1) = 2 W , 1 R = 2/9 x 1/8 x 3/7 = 6 / 504
Checking Dependency,
- P( x ∩ Y ) = 2/9 x 3/8 = 6/72
- P(x ) = 2/9
- P(y) = 3/9
- P(x∩y) ≠ P(x)P(y)
- Therefore x and y are not independent.
Joint probability distribution of (x,y) is found. Also x and y are not independent.
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