What is the probability of two out of three events happening? All events are independent.
$$\Pr(A) = \frac{9}{10}$$
$$\Pr(B) = \frac{9}{10}$$
$$\Pr(C) = \frac{6}{10}$$
What is the probability of at least two events happening?
I'd like to use negation, to negate the possibility that event no event happen plus the probability that only one happens.
$$D = \text{at least two events happen}$$
$$\Pr(D) = 1-\Pr(\text{none happens})-\Pr(\text{exactly one happens})$$
$$\Pr(D) = 1 - \left(\frac{1}{10}\cdot \frac{1}{10}\cdot\frac{4}{10}\right) - \left(\frac{1}{10}\cdot\frac{6}{10}\cdot\frac{1}{10} + \frac{9}{10}\cdot\frac{1}{10}\cdot\frac{4}{10} + \frac{9}{10}\cdot\frac{1}{10}\cdot\frac{4}{10}\right) = 0.918$$
The answer seems a little larger, I can't convince myself that I'm right. 

Can confirm that is the right answer, calculated it directly and obtained the same number.
