Another thought: Barnaby Jack was one of the top speakers and was to speak on a very controversial subject. I would guess that out of the 5000 speeches that were presented in your scenario only a few of them, maybe 2%, would contain information controversial enough that foul play would appear as a reasonable scenario to an outside observer (and this is being generous).

  Let F = foul play occurred in order to disrupt a conference,
    D = death of speaker one month before conference

  Let P(D) = (.02 deaths per year for 25-35 y.o) / (12 months in a year) = 0.0017
  Let P(F|D) = .001 (assuming 1 in 1000 chance foul play was involved given a death of a speaker at a conference)
    P(F) = P(F|^D) * P(^D) + P(F|D) * P(D) 
         = 0 * 0.98 + 0.001 * .02
         = 0.00002
  Let P(D|F) = 1 (chance of death if foul play is involved, assumed 100%)

  So Bayes theorem gives us:

  P(F|D) = P(D|F)*P(F)/P(D)
       = 1 * 0.00002 / 0.0017 
       = 0.018 (chance of foul play for a single speech given the speaker died 1 month before)

  Let P(C) = 0.02 (probability of a controversial speech)
  P(D) = .0017 (from above)

  Let P(C&D|F) = .5 (assuming there is a 50% chance the speech was controversial given foul play did occur, and death always occurs from foul play)
  P(C&D) = .02 * 0.0017 = 0.000034
  P(F) = 0.00002 (from above)

  P(F|C&D) = P(C&D|F) * P(F) / P(C&D)
         = 0.5 * 0.00002 / .000034
         = around a 30% foul play was involved in Barnaby Jack's death

edit: removed line "P(F|D) = 0.00058 (from above)" as pointed out by user 0003. End result didn't change, though.