The results of a monohybrid cross can be given as a probability, a ratio or a percentage. In a Punnett square for Bb × Bb there are four boxes, so each box has a probability of 1 in 4. This is 0.25, or 25%. The phenotype ratio is 3 dominant to 1 recessive.
Cystic fibrosis is caused by a recessive allele, f. Two carriers (Ff × Ff) have a 25% chance of a child with cystic fibrosis (ff), a 50% chance of a child who is a carrier (Ff) and a 25% chance of a child who is unaffected and not a carrier (FF). A cross between a carrier and a person who is homozygous dominant (Ff × FF) cannot produce an affected child, so the chance is 0%.
Probabilities show what is likely, not what must happen. To estimate how many offspring are expected to show a phenotype, multiply the probability by the number of offspring. For Bb × Bb the chance of the recessive phenotype is 25%, so out of 20 offspring we expect 5 to show it.
Pedigrees can be used to work out whether a trait is dominant or recessive. If two affected parents have an unaffected child, the trait must be dominant, and both parents must be heterozygous.