91Ó°ÊÓ

Referring to example 2.3.5,

The phenotype of an individual was observed. It was found that the individual is a dominant trait.

Referring to example 2.3.5, the events are defined as follows:

\({B_i}\)is the i^th genotype, \(\left( {i = 1,2,3,4,5,6} \right)\)

\(E\)is the event that the observed individual has the dominant trait.

Referring to example 2.3.5,

The probabilities of occurring the possible genotypes are,

\(\Pr \left( {{B_1}} \right) = \frac{1}{{16}}\),\(\Pr \left( {{B_2}} \right) = \frac{1}{4}\),\(\Pr \left( {{B_3}} \right) = \frac{1}{8}\),\(\Pr \left( {{B_4}} \right) = \frac{1}{4}\),\(\Pr \left( {{B_5}} \right) = \frac{1}{4}\),\(\Pr \left( {{B_6}} \right) = \frac{1}{{16}}\)

And the probability that the observed individual has the dominant trait, given the respective genotype occurs are,

\(\Pr \left( {E\left| {{B_1}} \right.} \right) = 1\),\(\Pr \left( {E\left| {{B_2}} \right.} \right) = 1\),\(\Pr \left( {E\left| {{B_3}} \right.} \right) = 1\),\(\Pr \left( {E\left| {{B_4}} \right.} \right) = \frac{3}{4}\),\(\Pr \left( {E\left| {{B_5}} \right.} \right) = \frac{1}{2}\),\(\Pr \left( {E\left| {{B_6}} \right.} \right) = 0\)

Using the probabilities stated above, the probability that the observed individual has the dominant trait is computed as follows,

\(\begin{aligned}{c}{\bf{Pr}}\left( {\bf{E}} \right){\bf{ = }}\sum\limits_{{\bf{i = 1}}}^{\bf{6}} {{\bf{Pr}}\left( {{{\bf{B}}_{\bf{i}}}} \right){\bf{Pr}}\left( {{\bf{E}}\left| {{{\bf{B}}_{\bf{i}}}} \right.} \right)} \\ = \frac{3}{4}\end{aligned}\)

The posterior probability that the parents have the genotypes of event\({B_i}\)is\(\Pr \left( {{B_i}\left| E \right.} \right)\).Also, the prior probability is represented by\(P\left( {{B_i}} \right)\).

Thus,

\({\bf{Pr}}\left( {{{\bf{B}}_{\bf{i}}}\left| {\bf{E}} \right.} \right){\bf{ = }}\frac{{{\bf{Pr}}\left( {{{\bf{B}}_{\bf{i}}}} \right){\bf{Pr}}\left( {{\bf{E}}\left| {{{\bf{B}}_{\bf{i}}}} \right.} \right)}}{{{\bf{Pr}}\left( {\bf{E}} \right)}}\)

Now, for the values of\(\left( {i = 1,2,3,4,5,6} \right)\), the probabilities are,

\(\begin{aligned}{c}\Pr \left( {{B_1}\left| E \right.} \right) = \frac{{\frac{1}{{16}} \times 1}}{{\frac{3}{4}}}\\ = \frac{1}{{12}}\\ > P\left( {{B_1}} \right)\end{aligned}\)

Similarly,

\(\begin{aligned}{c}\Pr \left( {{B_2}\left| E \right.} \right) = \frac{{\frac{1}{4} \times 1}}{{\frac{3}{4}}}\\ = \frac{1}{3}\\ > P\left( {{B_2}} \right)\end{aligned}\)

\(\begin{aligned}{c}\Pr \left( {{B_3}\left| E \right.} \right) = \frac{{\frac{1}{8} \times 1}}{{\frac{3}{4}}}\\ = \frac{1}{6}\\ > P\left( {{B_3}} \right)\end{aligned}\)

\(\begin{aligned}{c}\Pr \left( {{B_4}\left| E \right.} \right) = \frac{{\frac{1}{4} \times \frac{3}{4}}}{{\frac{3}{4}}}\\ = \frac{1}{4}\\ = P\left( {{B_4}} \right)\end{aligned}\)

\(\begin{aligned}{c}\Pr \left( {{B_5}\left| E \right.} \right) = \frac{{\frac{1}{4} \times \frac{1}{2}}}{{\frac{3}{4}}}\\ = \frac{1}{6}\\ < P\left( {{B_5}} \right)\end{aligned}\)

\(\begin{aligned}{c}\Pr \left( {{B_6}\left| E \right.} \right) = \frac{{\frac{1}{{16}} \times 0}}{{\frac{3}{4}}}\\ = 0\\ < P\left( {{B_6}} \right)\end{aligned}\)

Thus, on observation, it can be concluded that \({B_5}\) and \({B_6}\) have the posterior probabilities smaller than prior probabilities. That is,\(\Pr \left( {{B_5}\left| E \right.} \right) < \Pr \left( {{B_5}} \right)\) and \(\Pr \left( {{B_6}\left| E \right.} \right) < \Pr \left( {{B_6}} \right)\).