I have to solve the following problem: If a company's computer network has 2000 computers and 12 of them are broken,if we randomly select 10 computers,what are the odds of finding at least two broken ones?
I have used binomial distribution since we have repetition (a pc can be selected twice) which gives me a probability of P(X>=2)=0.00157 (https://homepage.divms.uiowa.edu/~mbognar/applets/bin.html)
Is this a correct way of tackling it? Thanks in advance
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
$$\mathbb{P}[X \ge 2]=1-\left(\frac{1988}{2000}\right)^{10}-10\left(\frac{12}{2000}\right)\left(\frac{1988}{2000}\right)^{9}\approx0.00157=0.15\%$$