<s>[INST] 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. 
 [/INST] Your approach works, and your answer is correct. 
It can also be calculated straight forwards as:
$\begin{align}
\mathsf P(AB\cup AC\cup BC)
& = \mathsf P(AB)+\mathsf P(A^cBC)+\mathsf P(AB^cC)
\\[1ex]
& = \mathsf P(A)\mathsf P(B) + \bigg(\mathsf P(A^c)\mathsf P(B)+\mathsf P(A)\mathsf P(B^c)\bigg)\mathsf P(C)
\\[0ex]
 &= \frac{9^2}{10^2} + \bigg(\frac{1\cdot 9}{10^2} + \frac{9\cdot 1}{10^2}\bigg)\frac 6{ 10}
\\[1ex]
& = \frac{918}{1000}
\end{align}$
</s>