Why is weighing random observations according to their probability from all distributions wrong? Is sampling all distributions n times and then talking out i numbers from each sample, where i is probability of that distribution * n, wrong? 
Suppose $$
0.3\!\times\mathcal{N}(0,1)\; + \;0.5\!\times\mathcal{N}(10,1)\; + \;0.2\!\times\mathcal{N}(3,.1)
$$ being my problem to sample 100 numbers. Should  I take 100 or say 1000 samples each of the three normal distributions and then take 30, 50 and 20 respectively, randomly out of them? 
Or going by another approach, Should I take 30, 50 and 20 random samples respectively, directly from the three distributions?
The correct algorithm seems to be:


*

*generate a number, say k  according to the probabilities of all distributions which corresponds to a particular k-th one. 

*generate a number from the above k-th distribution. 


Repeat for N numbers. See it  at sampling from a mixture of two Gamma distributions .
Are all these three approaches same? 
fwiw, I am using python and I am not familiar with R.  And the reason I am asking this question is this comment:

[..]The notation suggests that to sample, you need to sample all three normals and weigh the results by those coefficients which would obviously not be correct. [..]

here 
 A: The quote you refer to

The notation suggests that to sample, you need to sample all three
  normals and weigh the results by those coefficients which would
  obviously not be correct.

seems to misunderstand the notation. Mixture distribution of $m$ $f_k$ components and mixing weights $\pi_k$ is defined as
$$
f(x) = \sum_{k=1}^m \pi_k \; f_k(x)
$$
Weighting the distributions and weighting the values are not the same things. Moreover, we are looking at the probability of observing $x$ according to each of the $f_k$ distributions, not weighted sum of three different random variables
$$
\pi_X \, f_X(x) + \pi_Y \, f_Y(y) + \pi_Z \, f_Z(z)
$$
Drawing samples from three distributions and weighting them has nothing to do with mixture distribution. The notation does not suggest anything like this It is the opposite, we are summing the components because they are mutually exclusive.
So the correct way of thinking about mixture is that you first randomly choose the $k$-th component with probability $\pi_k$, and then draw a sample from this component according to the distribution $f_k$. Same happens in the algorithm for sampling from mixture distribution.
As about your proposed algorithm, it is not equivalent to the proper algorithm. If you needed to simulate 100 draws using a fair coin, you would not take 50 heads and 50 tails and shuffle them, this would not be a valid sample. If the probability of drawing heads is 0.5 this does not mean that in sample of size $n$ you would observe $n\times 0.5$ heads. It means that with $n$ large enough you would see approximately that many heads. Same applies to mixtures, you need to draw the components randomly.
Sorry but I don't follow what you mean by your first algorithm, where you want to draw 1000 samples to obtain 100 samples from the mixture.
