This is a pedigree for autosomal recessive disease albinism (aa). What is the probability of II – 1 being homozygous normal?

The correct option is A 1/3
In the pedigree given, both the individuals of I generation are normal, but one offspring is affected out of three. Hence the genotype of both individuals of I generation is Aa, i.e., both parents carry a copy of 'a' gene.
All the possible genotype of the II generation are:

$\begin{array}{ccc}& \text{A}& \text{a}\\ \text{A}& \text{AA}& \text{Aa}\\ \text{a}& \text{Aa}& \text{aa}\end{array}$

Therefore, there are 3 possible genotypes: AA (homozygous normal), Aa (heterozygous normal), aa (affected with albinism). We can see from the pedigree that the person (II-1) is unaffected, hence she does not have aa gene and hence we can rule out the aa instance. So, out of the remaining three intances i.e., AA, Aa and Aa, only in 1 instance the person can be homozygous normal i.e., have AA gene.
Hence, the probability of an individual being homozygous normal will be 1 (AA) out of 3 (AA, Aa and Aa)$=\frac{1}{3}$.
So, the correct option is a.