Study the given pedigree chart showing the inheritance of a genetic...

From the pedigree, unaffected parents I-1 and I-2 have an affected daughter II-1. This indicates that the trait is autosomal recessive.

Let the recessive allele be '$a$' and the dominant allele be '$A$'.

Since I-1 and I-2 are unaffected but have an affected child ($aa$), both must be heterozygous ($Aa$).

Individual II-2 is an unaffected child of I-1 and I-2. The probability that II-2 is a carrier ($Aa$) is $\dfrac{2}{3}$ (since the possible genotypes for an unaffected child are $AA$ and $Aa$ in a $1:2$ ratio).

Individual I-3 is affected ($aa$), which means he must pass the '$a$' allele to all his children. Since his daughter II-3 is unaffected, she must be heterozygous ($Aa$) with a probability of $1$.

For the child III-1 to be affected ($aa$), both parents II-2 and II-3 must be carriers ($Aa$) and both must pass on the '$a$' allele.

Probability that III-1 is affected = Probability(II-2 is $Aa$) $\times$ Probability(II-3 is $Aa$) $\times$ Probability(two heterozygotes have an $aa$ child)

Probability = $\dfrac{2}{3} \times 1 \times \dfrac{1}{4} = \dfrac{1}{6}$.

Answer: $\dfrac{1}{6}$