关于一定概率随机函数的算法及实现

Question

有以下需求：
  1、以p的概率产生0，以1-p的概率产生1
  2、有具体的算法实现分不够的话可以另开贴

Answer 1

    Random rand = new Random();
        static void Main(string[] args)
        {
            Random rand = new Random();
            double dou = 0.8888;
            Thread thread = new Thread(new ThreadStart(() =>
                {
                    int pCount = 0;
                    int i = 0;
                    while (i < 1000)
                    {
                        i++;
                        if (rand.NextDouble() < dou)
                        {
                            pCount++;
                            Console.WriteLine(i + ":  p");
                        }
                        else
                        {
                            Console.WriteLine(i + ":  1-p");
                        }
                    }
                    Console.WriteLine("p  " + pCount + " times.");
                    Console.WriteLine("1-p " + (1000 - pCount) + " times");
                    Console.WriteLine("set 88.88%，actual " + (pCount / 10.000000) + "%");
                }));
            thread.Start();
            Console.Read();
        }

Answer 2

用Random产生一个0~1的随机数，小于0.0987就是中奖。