Determine the genotype of the albino child.

Since albinism is a recessive trait, an albino individual must have two copies of the recessive allele.
Albino child genotype: $aa$

Both parents have normal skin colour (dominant phenotype) but produced an albino child (aa). This means each parent must have contributed an 'a' allele to the albino child. Therefore, both parents must be heterozygous carriers.
Man's genotype: $Aa$
Woman's genotype: $Aa$

When two heterozygous parents (Aa x Aa) cross, the possible genotypes of their offspring are:

The two children with normal skin colour could have genotypes AA or Aa.

Probable genotypes:

• Parents: Man (Aa), Woman (Aa)
• Children: One albino (aa), two normal skin (AA or Aa)

Pedigree Diagram (text description):
Generation I:
Male (normal skin, carrier) -- Female (normal skin, carrier)
Genotypes: Aa -- Aa

Generation II:
Offspring:

1. Child (normal skin): Genotype AA or Aa
2. Child (normal skin): Genotype AA or Aa
3. Child (albino): Genotype aa

The cross Aa x Aa produces offspring with genotypes: 1 AA : 2 Aa : 1 aa.
Among the offspring with normal skin colour (AA and Aa), the ratio is 1 AA : 2 Aa.
So, the probability that a normal-skinned boy is AA is $\frac{1}{3}$.
The probability that a normal-skinned boy is Aa is $\frac{2}{3}$.

The girl is a carrier for the albino gene, so her genotype is $Aa$.

We consider two scenarios for the boy's genotype:

Scenario 1: The boy is AA (probability $\frac{1}{3}$).
Cross: $AA \times Aa$
Offspring genotypes: $1/2 AA, 1/2 Aa$
All offspring (100%) will have normal skin colour.
Probability of normal skin in this scenario = 1.
Contribution to total probability = $\frac{1}{3} \times 1 = \frac{1}{3}$.

Scenario 2: The boy is Aa (probability $\frac{2}{3}$).
Cross: $Aa \times Aa$
Offspring genotypes: $1/4 AA, 1/2 Aa, 1/4 aa$
Offspring with normal skin colour (AA or Aa) = $\frac{1}{4} + \frac{1}{2} = \frac{3}{4}$.
Probability of normal skin in this scenario = $\frac{3}{4}$.
Contribution to total probability = $\frac{2}{3} \times \frac{3}{4} = \frac{6}{12} = \frac{1}{2}$.

Total probability of the first child having normal skin colour = $\frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}$.

The probability that their first child will have normal skin colour is $\boxed{\frac{5}{6}}$.