Taking as example the crossing of AaBbCc with aaBBCc, for each considered pair of allele it is possible to verify which genotypes it can form (as in an independent analysis) and in which proportion. AA x aa: Aa, aa (1:1). Bb x BB: BB, Bb (1:1). Cc x Cc: CC, Cc,
cc (1:2:1). The genotype to which the probability is to be determined is for example aaBbcc. For each pair of this genotype the formation probability is determined: to aa, 0.5; to Bb, 0.5; to cc, 0.25. The final result is obtained by multiplication of these partial probabilities, 0.5 x 0.5 x, 0.5, resulting 0.0625.