Consider the events Ea, Eb, and Ec indicating that the extravert is Albert, Bob, and Carl respectively. All of these events have an equal probability of 1/3 to begin with.
Dr. Fraud initially choosing Albert, Bob, or Carl to be the extravert is described the events Xa, Xb, or Xc respectively. As Dr. Fraud's initial choice is independent of who the extravert actually is, the conditional probabilities describing who the extravert is given that he has chosen one is still 1/3 for all scenarios. Dr. Fraud meeting with Albert, Bob, or Carl is indicated by the events Ma, Mb, or Mc respectively.
P(Ea|Mb,Xc)
= P(Mb|Ea,Xc) · P(Ea|Xc) ÷ P(Mb|Xc)
= P(Mb|Ea,Xc) · P(Ea|Xc) ÷ [ P(Mb|Ea,Xc) · P(Ea|Xc) + P(Mb|Eb,Xc) · P(Eb|Xc) + P(Mb|Ec,Xc) · P(Ec|Xc) ]
= P(Mb|Ea,Xc) ÷ [ P(Mb|Ea,Xc) + P(Mb|Eb,Xc) + P(Mb|Ec,Xc) ]
= (1) ÷ [ (1) + (0) + (1/2) ] = 2/3
P(Ec|Mb,Xc)
= P(Mb|Ec,Xc) · P(Ec|Xc) ÷ P(Mb|Xc)
= P(Mb|Ec,Xc) · P(Ec|Xc) ÷ [ P(Mb|Ea,Xc) · P(Ea|Xc) + P(Mb|Eb,Xc) · P(Eb|Xc) + P(Mb|Ec,Xc) · P(Ec|Xc) ]
= P(Mb|Ec,Xc) ÷ [ P(Mb|Ea,Xc) + P(Mb|Eb,Xc) + P(Mb|Ec,Xc) ]
= (1/2) ÷ [ (1) + (0) + (1/2) ] = 1/3