Probability of event with repetition [closed]
Actually your sampling is without replacement thus the correct answer is
$$\mathbb{P}[X\ge 2]=1-\mathbb{P}[X=0]-\mathbb{P}[X =1]=1-\frac{\binom{1988}{10}+\binom{12}{1}\binom{1988}{9}}{\binom{2000}{10}}\approx0.001447=0.14\%$$
To simplify the calculations, in this example you can use binomial finding, without a lot of calculations
$$\mathbb{P}[X \ge 2]=1-\left(\frac{1988}{2000}\right)^{10}-10\left(\frac{12}{2000}\right)\left(\frac{1988}{2000}\right)^{9}\approx0.00157=0.15\%$$