How is an RT-PCR representative of the RNA contents in a cell?
It isn’t really clear exactly what the questioner is asking, but…

...one way of thinking about this is to ask how likely is it that a low-abundance RNA will be absent from the sample taken from the preparation of RNA. This would make the sample non-representative at the qualitative level.

The question states that 12,000 ng of RNA is isolated. Since nothing else is specified I assume this is total RNA. A rough estimate of total RNA per eukaryotic cell is 20 pg.

This means that the preparation came from 6 x 105 cells. (I’m assuming 100% yield, but this won’t really affect the conclusion.)

Now assume that a low-abundance mRNA is present at 10 copies per cell. The total RNA preparation therefore contains 6 x 106 copies of that low-abundance mRNA.

We take a sample of 1,000 ng from the total 12,000 ng. In other words we take 1/12th of the preparation of RNA. If that RNA preparation contains 106 copies of the low abundance mRNA how likely is it that we happen to take a sample which contains no copies of that low abundance mRNA? Here my maths fails me, but intuitively I say that the probability is very very small.

Now, if there are 1,000 different low-abundance mRNAs in the cell the probability of missing each of them is very very small, but the overall probability of missing one of them is 1000 x very very small, so maybe very small.

I'll continue thinking about how to calculate the value of very very small, but if anyone wants to help out - thanks in advance!