Ok I think I have it now :))
It is noted that 8% of students are left handed. If 20 (TWENTY) students are randomly selected,
P( no one in a class of 20 is left handed) = 0.92^20 = 0.18869
P( one or more in a class of 20 is left handed = 1-0.18869 = 0.81131
(b) If 50 (FIFTY) classes of 20 (TWENTY) students are randomly selected, what is the probability that 10 (TEN) classes have no left-handed students?
P(10 classes of 20 students have no left-handed students)
= 50C10 * 0.18869^10 * 0.81131^40
nCr(50,10) * 0.18869^10 * 0.81131^40 = 0.1369710474047608673359021944148508674768838
The probability is 0.137