<s>[INST] Probability of balls in the box In a box there are 12 balls; 4 defective, 8 not defective.
What is the probability that when 3 balls are drawn, at least two of them are defective.
I know the answer is $$\frac{{4 \choose 2}{8 \choose 1} + {4 \choose 3}}{{12 \choose 3}}$$
But why isn't the answer also $$\frac{{4 \choose 2}{10 \choose 1}}{{12 \choose 3}} $$?
Because I choose 2 balls from the 4 defective, and 1 from the remaining 10 (it will either be defective or not).
 [/INST] Another way is to look at the possible sequences. Let's say $D$ is a defect ball and $G$ is a nondefect (good ball). If you sample 3 balls without replacement the event 'to get at least 2D' is $\{DDD,GDD,DGD,DDG\}$. The probability of the first event is 
$P(DDD)= \frac{4 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10}$
It seems that the probability of the other 3 events has to be different, but it is actually not:
$P(GDD)=P(DGD)=P(DDG)=\frac{4 \cdot 3 \cdot 8}{12 \cdot 11 \cdot 10}$ because the denominator does not change and the numerator is the product of the same 3 numbers (and multiplication is commutative), therefore:
$P(D \geq 2)=P(DDD)+P(GDD)+P(DGD)+P(DDG)=\frac{4 \cdot 3 \cdot 2}{12 \cdot 11 \cdot 10}+3 \cdot \frac{4 \cdot 3 \cdot 8}{12 \cdot 11 \cdot 10}=.2364$
Control:$1-P(D<2)=1-\Big( \frac{8 \cdot 7 \cdot 6}{12 \cdot 11 \cdot 10}+3 \cdot \frac{8 \cdot 7 \cdot 4}{12 \cdot 11 \cdot 10} \Big)=.2364$
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