Examples using population allele frequencies.

Example 1: Let's say that among pea plants, the frequency of the red flower-color allele is 0.7, and the frequency for the white flower-color allele is 0.3. What is the probability that a pea plant will be red-red; red-white; white-white?
Although we are working with allele frequencies on a population-wide basis instead of just within a "family", the same rules apply. These pea plants will get two flower-color alleles, one from each parent. Thus, there is one red-red combination, two red-white combinations, and one white-white combination. Therefore the chance that a plant will be red-red is 0.7*0.7 = 0.49. The chance that a plant will be red-white is 2(0.3*0.7) = 0.42. And, the chance that a plant will be white-white is 0.3*0.3 = 0.09.

Example 2: Let's say that in a population, there is a common recessive albino trait. The normal allele has a frequency of 3/4, and the albino allele has a frequency of 1/4. We'll call them N and D respectively.
A woman who is ND marries a man with the normal phenotype but unknown genotype. What is the chance that the man will pass on an N allele? A D allele? That they will have a child who is NN; ND; DD?
First of all, we know that the that the man is not DD (P(DD|normal)=0). We need the conditional probabilities that the man is DN or NN, given he has normal phenotype.
P(albino) = P(DD) = (1/4)*(1/4), so P(normal)= 1- P(DD) = 15/16.
P(NN | normal) = P(NN and normal)/P(normal) = P(NN)/P(normal) = ((3/4)*(3/4))/(15/16) = 9/15 = 3/5.
P(ND | normal) = P( ND and normal)/P(normal) = (2*3/4*1/4)/(15/16) = 6/15.
(Note 9/15 + 6/15 = 1; if the man is normal he must be NN or ND)

The chance the man passes on an N is 1/2 if he is ND and 1 if he is NN, so overall is (1/2)*(6/15) + 1*(9/15) = 12/15 = 4/5.
The chance that the man passes on a D is 1/2 if he is ND and 0 if he is DD, so overall is 1/2*6/15 + 0 =3/15 = 1/5.
(Note 4/5 + 1/5 = 1; the man must pass on an N or a D)

Now, we know the mom is a carrier (ND), so her chance of passing on the D allele is 1/2, and her chance of passing on the N allele is 1/2.
The chance of these two people having a child who is NN is 1/2*4/5 = 2/5.
The chance of a child who is DD is 1/2*1/5 = 1/10.
The chance of a child who is ND is 1/2*4/5 + 1/2*1/5 = 1/2 (why?)
Check: 2/5 + 1/10 + 1/2 = 1

Example 3: Now, using the same trait and allele frequencies as the example before, let's say that both parents have normal phenotypes with unknown genotypes. What is the chance that they have a child who is NN? ND? DD?
We already know from before that under these circumstances, each parent has a 4/5 chance of passing on the N allele, and a 1/5 chance of passing on the D allele. (see computations for father in previous example) Therefore:
the chance that the child is NN is 4/5*4/5 = 16/25
the chance that the child is DD is 1/5*1/5 = 1/25
the chance that the child is ND is 2*4/5*1/5 = 8/25
Check 16/25 + 1/25 + 8/25 = 1