Thanks for the compliment. This is a good problem. The probability that exactly one employee will be in the bottom 5% is 5*(.05)*(.95)4 = 0.203627. Given that one employee is in the lowest 5% the probability of exactly one in the next 20% is 4*(.2/.95)*(.75/.95)3 = 0.414361. Given these two underachievers the probability of exactly one in the next 50% out of the remaining 75% is 3*(.5/.75)*(.25/.75)2 = 0.222222. The probability that one of the remaining two falls in the lower 20% of 25% is 2*(.2/.25)*(.05/.25) = 0.32. Taking the product of all these probabilities we get 0.006, or 3/5 of 1%.