Question 1051563
12 bulbs:  9 working, 3 def
p(working = 9/12 = 3/4
testing 3, each one in turn, probability of finding 2 working
(a) among the three bulbs tested (one def)
P(2 out of 3) = (3/12)(9/11)(8/10) + (9/12)(3/11)(8/10) + (9/12)(8/11)(3/10)
(b) when three bulbs have been tested, but not before
P = 1 -P(all def) - P(2 def)
p = {{{1 - (3C3)(9C0)/(12C3) - (3C2)(9C1)/(12C3)}}}