a box contians 12 bulbs with 3 defective ones .if two bulbs are dramn from the box together,what id the probability

Answer 1

incomplete, I will assume it will end like ...
"if two bulbs are drawn from the box together, what is the probability that one is good and the other defective"
prob(defective) = 3/12 = 1/4
prob(good) = 3/4

prob(as assumed) = C(2,1)(1/4)(3/4) = 3/8

btw, due to the few number of cases, you could just do each possibility
GG = (3/4)(3/4) = 9/16
GB = (3/4)(1/4) = 3/16
BG = (1/4)(3/4) = 3/16
BB = (1/4)(1/4) = 1/16

notice that they add up to 1
and that the scenario I assumed would be 3/16+3/16 = 3/8