This seems pretty low, can't we apply Bayes' theorem here?

  let F = transaction being fraud
  let D = detected as fraud
  
  P(F) = 492/284807 = .17%
  P(D|F) = 80%
  P(~F) = 1-P(F) = 99.83%
  P(D|~F) = 13%
  
  P(F|D) = (P(D|F) * P(F)) / (P(F) * P(D|F) + P(~F) * P(D|~F))
  P(F|D) = ((.8) * (.0017)) / ((.0017) * (.8) + (.9983) * (.13))
  P(F|D) = 0.01 = 1% chance of the transaction being fraud given a positive detection

although to be fair, they're not stating to only use this method by itself