In humans wavy hair(W) is dominant over straight(w). Dark hair(D) dominates red(d).A wavy dark haired male(heterozygous dominant for both traits) married a wavy red haired woman(homozygous recessive for both traits).What is the probability of having a wavy red haired child ?

But it seems to be a mistake in the statement .."married a wavy red haired woman(homozygous recessive for both traits)". If the woman has wavy hair, the trait can't be homozygous recessive, although she certainly is homozygous recessive for the second trait because is red hair.

Step-by-step explanation:

As wavy hair is dominant over straight hair, it means only one allele is necessary to be visible: WW (homozygous), and Ww (heterozygous) produce the wavy hair trait. An homozygous recessive (ww) doesn't show the wavy trait character; he or she would have straight hair

The woman, if she's homozygous and wavy hair, would be (WW). As she is homozygous for red hair, she has red hair.

If the statement "A wavy dark haired male(heterozygous dominant for both traits) married a wavy red haired woman(homozygous recessive for both traits)" is correct, meaning that she is not wavy hair but "STRAIGHT", the probability of having a wavy red hair changes, being only 1/4

Calculation can be performed with Punnett squares

For example for the wavy dark hair male (heterozygous for both traits), the genotype is WwDd and the possible allele combinations would be: WD, Wd, wD, and wd

For the woman, if she's wavy red hair (homozygous), the genotype would be WWdd and the possible allele combinations would be only Wd.

Then you need cross the male allele combinations against Wd

If the woman is straight red hair (homozygous for both traits), the genotype would be wwdd and the possible allele combinations would be only wd.

Then you need cross the male allele combinations against wd, and obtain the proportions produced from the cross

An example of Punnett square below