## Exclusion Probabilities

Complex DNA mixtures can be conservatively interpreted using the combined probability of exclusion (CPE) (Devlin 1992, DAB 2000, Ladd et al. 2001). The probability of exclusion provides an estimate of the portion of the population that has a genotype composed of at least one allele not observed in the mixed profile (DAB 2000). If evidence includes three alleles at a locus (A1, A2, A3), then:

p = A1 + A2 + A3 and q = 1 — p so PE = 2 pq + q2.

The probability of exclusion (PE) at each locus is first determined and then PEs from multiple loci are combined through the equation listed below:

Combined PE = 1 — [(1 — PE1) x (1 — PE2) x ... x (1 — PEn)]

In the example shown in Figure 22.2, determination of CPE would involve calculation of the frequency of genotypes that do not possess the alleles 11, 12 or 14 present in the evidentiary mixed profile. The CPE approach avoids the potential pitfalls associated with extrapolating the genotypes of mixture contributors (e.g., overlooking an allele because it is in a stutter position). No prior knowledge or assumptions regarding the number of possible contributors to the mixture are needed and results can still be reported without knowledge of a known profile, such as the victim's profile.

While calculating the combined probability of exclusion is not as powerful of a technique as the likelihood ratio method that is discussed in the next section, a major advantage with the CPE approach is that the number of contributors to the crime scene DNA profile does not need to be taken into account. Simply all other alleles not observed in the stain are considered. This approach is rather conservative because an individual can be excluded if he has any allele at any locus that is not detected in the stain.

 D13S317 Possible Alleles Let p = A1 + A2 + A3 (with q =1-p) A Freq (p) PE = 2pq + q2 15 J 0.00828* Evidence 14 l ] 0.04801 1 1 A1 1 p = p(A11) + p(A12) + p(A14) = 0.33940 + 0.24834 + 0.04801 13 - 0.12417 = 0.63575 12 l J 0.24834 1-1 A2 q = 1 p = 1 0.63575 = 0.36425 11 ; : 0.33940 1 1 A3 PE = 2(0.63575)(0.36425) + (0.36425)2 10 ; : 0.05132 = 0.46314 + 0.13268 = 0.59582 9 0.07450 8 0.11258 Fom Table 20.2

Figure 22.2

Calculation of probability of exclusion with a mixture example.

Figure 22.2

Calculation of probability of exclusion with a mixture example.