If a female is a carrier for colour blindness, marries a male which is normal, then in the offsprings this disease may be seen in ?

50% sons diseased and 50% daughters (carrier).

Option 2 is correct . 50% sons diseased  and 50% daughters (carrier).

Inheritance of color blindness follows an X-linked recessive pattern, where the gene responsible for color blindness is located on the X chromosome. Females have two X chromosomes (XX), while males have one X and one Y chromosome (XY). If the mother is a carrier (possesses one normal X and one X with the color blindness gene) and the father does not carry the gene, the following inheritance pattern may occur:

Daughters: Each daughter has a 50% chance of inheriting the X chromosome carrying the color blindness gene from the mother. Therefore, they become carriers like their mother but usually do not express the trait because they also inherit a normal X chromosome from their father. If They carry other normal X chromosome from mother and father then 50% will be normal daughters.

Sons: Sons inherit the Y chromosome from their father, not the X chromosome carrying the color blindness gene. As a result, they do not inherit the disease because they only have one X chromosome (from the mother) and one Y chromosome (from the father). But if they carry the X chromosome carrying the color blindness gene from the mother and Y from father then they will become diseased.

Therefore, in this scenario, daughters have a 50% chance of becoming carriers of the color blindness gene,and 50% of becoming normal, while sons will have 50 % chance of becoming diseased and 50% will be normal if they carry normal X chromosome from their mother.