Abox contains 12 transistors, 3 of which are defective. if 3 are selected at random, find the probability that
a all are defective
b none are defective

I guess your are selecting them without replacement, if so:

a) P(picking one defective) = 3/12
P(picking a 2nd defective) = 2/11
P(picking a 3rd defective) = 1/10

P(1 and 2 and 3 defective) = 3/12 x 2/11 x 1/10 = 1/220 = 0.0045

b) None defective: P( Non Defective) = 9/12 (follow Same logic):
9/12 x 8/11 x 7/10 = 21/55 = 0.38

Second method using combination:

a) ³C₃ / ¹²C₃ = 1/220 0.0045
B) ⁹C3 / ¹²C³ = 21/55 = 0.38

The answer would be -3.9> -4.5