Given that:
Locus 1:
Frequency of Allele 1 = p1 = 0.5
Frequency of Allele 2 = q1 = 0.1
Locus 2:
Frequency of Allele 1 = p2 = 0.7
Frequency of Allele 2 = q2 = 0.2
Locus 1:
Frequency of Allele 1 = p3 = 0.8
Therefore,
Genotypic frequency of Locus 1 with Allele 1 and Allele 2 =
2p1q1 = 2 * 0.5 * 0.1 = 0.1
Genotypic frequency of Locus 2 with Allele 1 and Allele 2 =
2p2q2 = 2 * 0.7 * 0.2 = 0.28
Genotypic frequency of Locus 3 with Allele 1 =
(p1)2 = 0.8 * 0.8 = 0.64
Therefore, frequency of co-occurence of these three Genotypes = 0.1 * 0.28 * 0.64 = 0.01792
Therefore the probability that a random individual from the population would match the evidentiary sample is 0.01792 or 1.792%
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Here is an example using 3 CODIS STR results. DNA evidence was found at a crime scene. Due to the poor quality...