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An urn contains 4 white and 3 red balls. Let X bethe number of red balls in a random draw of three balls. Find the mean andvariance of X. |
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Answer» There are 4 cases here. `Case 1:` When no red ball is drawn. All drawn balls are white.Then, `P(X=0) = (C(4,3)**C(3,0))/(C(7,3)) = (4**1)/((7**6**5)/(3**2**1)) = 4/35` `Case 2:` When one red ball is drawn. Other two drawn balls are white.Then, `P(X=1) = (C(4,2)**C(3,1))/(C(7,3)) = (6**3)/35 = 18/35` `Case 3:` When two red balls are drawn.Third ball is white.Then, `P(X=2) = (C(4,1)**C(3,2))/(C(7,3)) = (4**3)/35 = 12/35` `Case 4:` When no white ball is drawn. All drawn balls are red.Then, `P(X=3) = (C(4,0)**C(3,3))/(C(7,3)) = (1**1)/35 = 1/35` `:.` Mean, `E(X) = sum x_iP(x=x_i) = 0**4/35+1**18/35+2**12/35+3**1/35` `=>E(X) = 18/35+24/35+3/35 = 45/35 = 9/7` Now, Variance`(sigma^2) = E(X^2) - (E(X))^2` `=>sigma^2 = sumx_i^2P(x=x_i) - (E(X))^2` `=>sigma^2 = 0**4/35+1**18/35+4**12/35+9**1/35-(9/7)^2` `=>sigma^2 = 0+18/35+48/35+9/35-81/49 = 75/35-81/49` `=>sigma^2 = 15/7-81/49 = (105-81)/49 = 24/49` So, Mean is `9/7` and variance is `24/49`. |
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