This is specified by NCBI (see link).  

If MAF is 0.12 in your population, it means that 12% of your population will carry at least one 2nd most common allele (heterozygous or homozygous mutant). In other words, MAF is your probability for an individual to be a carrier (regardless whether on one or both chromosomes). 

It may also be instructive to calculate the carrier frequency of a simple recessive based on the percentage of phenotypically affected individuals, which assuming HWE and two alleles works as follows:

a + b = 1

a^2 + 2ab +b^2 = 1

Where a represents the dominant and b the recessive allele's frequency within the gene pool. Now let's assume for the sake of argument that the observed frequency of the recessive phenotype is 1% or 0.01, which equals b^2. Hence, b is sqrt(0.01), or 0.1, which is your MAF.

However, this does not equal the probability of any individual being a carrier, which is actually expressed as 2ab + b^2, which in our example would be 2*0.9*0.1 + 0.1^2 = 0.19 (or 0.18 if you're not interested in the homozygous individuals).

Thank you both for taking time to answer my question.

The link (Jan's comment) doesn't seem to work! I am very curious to see what NCBI is writing about this, as I always thought that MAF is at "chromosome level" and therefore does not tell how the alleles are distributed in individuals in the population. I myself also use the Hardy-Weinberg equilibirum equation as Silvan suggests, so I am glad to learn that I am not the only one doing that.

Thanks,

Tina