We can use the following Punnett square to do the crossing and
get the results. Let aa is the recessive homozygote and Aa is
heterozygote.
                             Â
l   a
  l  a  l
                           Â
A l
Aa l Aa l
                           Â
a l aa
l aa l
Now, we see that of the above four offsprings of generation F1,
two are heterozygous. Therefore, percent chance of getting a
heterozygote in this cross is 50%.
Thanks!