For example, the chance of any person in the Caucasian population having cystic fibrosis is 0.025*0.025 since that is the allele frequency in the population and two mutant alleles are needed for the disease phenotype. Likewise, the chance of someone being a carrier for cystic fibrosis is 2*(0.025*0.975) since, by definition, a carrier has one normal and one mutant allele. The factor 2 here is because a person could get the normal allele from either parent, but must then get the mutant allele from the other parent. For a person to be unaffected and not a carrier, the probability is 0.975*0.975.
By contrast, if you were to look at a family where you knew that both the parents were carriers for CF, you would know that the probability of them having a child who is normal and not a carrier is 0.5*0.5. The chance of them having a child who is a carrier is 2(0.5*0.5). The chance of them having a child with CF is 0.5*0.5. If you do not know the genotypes of the parents, then you have to use population allele frequencies and what you can determine about the genotypes of the parents based on their phenotypes to calculate the probability of them having a child with CF.