In females, X-linked recessive disorder is expressed only when the mutated gene is present in both the X chromosomes whereas in males presence of only one copy of mutated gene is sufficient to cause the disease as they have only one X chromosome.
Now the given cross will results in the production of:
Two daughters- one is homozygous dominant () and other is heterozygous dominant
Two sons- one carries dominant gene () and other carries recessive or mutated gene ().
From the test, it is clear that only one son out of 4 offspring carries the disease.
Hence, the probability will come out to be 1/4 = 0.25 or 25 percent.