An urn contains a total of N balls, some of which are red and the rest are black. A sample of 3 balls is drawn at random without replacement. Let X denote the number of red balls in the sample. If the mean of X is 1 and the variance of X is 3 5 , then the number of black balls in the urn is _________

Step-by-step solution

Let the total number of balls be N and the number of red balls be K . The number of balls drawn is n = 3 . For a sample drawn without replacement, X follows a hypergeometric distribution. The mean of X is given by E(X) = n K N . Given E(X) = 1 , we have 3 K N = 1 K N = 1 3 . The variance of X is given by Var(X) = n K N (1 - K N ) N-n N-1 . Given Var(X) = 3 5 , we substitute n=3 and K N = 1 3 : 3 ( 1 3 ) (1 - 1 3 ) N-3 N-1 = 3 5 1 2 3 N-3 N-1 = 3 5 2(N-3) 3(N-1) = 3 5 Cross-multiplying yields: 10(N-3) = 9(N-1) 10N -