Two people took turns tossing a fair die until one of them tossed a 6. Person A tossed first, B second. A third, and so on. Given that perosn B threw the first 6, what is the probability that B obtained the first 6 on her second toss (that is on, on the fourth toss overall)? Solution Geometric pmf -> x = n, p(1-p)^n P[6] = probability of rolling a 6 = 1/6 P[x = 4] = (1/6)(5/6)^3 = 125/1296 = .09645.