Answer:
1/4
Explanation:
Let achondroplastic dwarfism be controlled by gene A
Let red green color blindness be controlled by gene Xc
Man : He is dwarf so he will be either AA or Aa since its a dominant trait. His father had normal height so he must be aa. He will pass on one "a" allele to his son. Hence, the man's genotype is Aa. He has normal vision and colorblindness is recessive X-linked trait so his total genotype is AaXCY
Woman : She has normal height so her genotype is aa. She is colorblind hence her total genotype will be aaXcXc.
When,
AaXCY X aaXcXc :
If we check the crosses individually:
A a
a Aa aa
a Aa aa
XC Y
Xc XCXc XcY
Xc XCXc XcY
All the sons will be colorblind (XcY) and half of the children will have normal height (aa). Out of these children half will be sons so total probability of having a colourblind son with normal height will be :
1 * 1/2 * 1/2 = 1/4