Huntington's disease is a dominant condition with late age of onset in humans. if one parent has the disease, what is the probability that his or her child will have the disease

For the sake of calculation, let us assume that the diseased parent is homozygous for the Huntington’s disease. Homozygous meaning that the diseased parent’s gene carries two copies of the same allele.

Since only 1 of the 2 parents has the disease, then the probability of contracting it is ½.

1/2

We have to do both the table?? And we have to do bar graph?? Can you say please Because the question you mention is not in detail.. ☺️