The answer is 3/8
Let T be a dominant allele for tongue rolling and let t be a recessive allele.
Tongue-rolling is a dominant trait. That means that an individual with at least one dominant allele will have the trait:
TT - dominant homozygous individuals able to roll tongue
Tt - heterozygous individuals able to roll tongue
tt - recessive homozygous individuals able to roll tongue
A heterozygous man (Tt) and recessive homozygous female (tt) will have 2 of 4 children that are able to roll tongue (Tt). Take a look at the attached image: 2 of 4 children (marked yellow on the image) are able to roll tongue (Tt) and 2 of for children are not (tt).
The probability that a child will be able to roll his/her tongue is 2/4 = 1/2
Now, let's see what happens with color blindness as an X-linked recessive trait. Let \(X_V\) be a dominant allele and \(X_v\) be a recessive allele. The genotypes will be as following:
\(X_VX_V\) - a female with normal vision
\(X_VX_v\) - a heterozygous female with normal vision
\(X_vX_v\) - a female with colorblindness
\(X_VY\) - a male with normal vision
\(X_vY\) - a male with colorblindness
Take look at the attached image. A man with normal vision (\(X_VY\)) and a woman heterozygous for colorblindness (\(X_VX_v\)) will have 3 of 4 children that have normal vision (marked yellow on the image) and 1 of 4 children with color blindness.
The probability that the child has normal vision is 3/4.
Now, we have two probabilities:
The probability that a child will be able to roll his/her tongue is 1/2.
The probability that the child has normal vision is 3/4.
Since we want both of these traits to occur together, we will just multiply their possibilities:
1/2 * 3/4 = 3/8
