I like this probability test. There are i white balls and n-i black balls with 0 le i lq n. The possible combinations are n, so that the number of white balls is n(n+1)/2 and each combination has 2/n(n+1) probability. Now suppose that we fix the number of white balls to k. The probability to extract the first white is 2k/n(n+1). The probability of extracting the second white is (k-1)/(n-1), since there are n-1 balls left. The two events are independent, so that the conditional probability is the product of 2k(k-1)/n(n-1)(n+1). The overall probability of all the combination is \sum_k=0..n 2k(k-1)/n(n-1)(n+1)

I like this probability test. There are i white balls and n-i black balls with 0 le i lq n. The possible combinations are n, so that the number of white balls is n(n+1)/2 and each combination has 2/n(n+1) probability. Now suppose that we fix the number of white balls to k. The probability to extract the first white is 2k/n(n+1). The probability of extracting the second white is (k-1)/(n-1), since there are n-1 balls left. The two events are independent, so that the conditional probability is the product of 2k(k-1)/n(n-1)(n+1). The overall probability of all the combination is \sum_k=0..n 2k(k-1)/n(n-1)(n+1)

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