What are the chances: at least once, expected count, the tail

atLeastOnce[p_, n_] := 1 - (1 - p)^n

N[atLeastOnce[1/100, 50]]

0.3949939328624633

expectedHits = 50 (1/100)

1/2

atLeastTwice = N[1 - (1 - 1/100)^50 - 50 (1/100) (1 - 1/100)^49]

0.08943531309603069

Plot[atLeastOnce[1/100, n], {n, 0, 300}]

-Graphics-