Question 657829
There are {{{(matrix(2,1,5,2)) = 5!/(2!*(5-2)!) = (5*4)/(2*1) = 10}}} ways to choose 2 defective bulbs from the five defective bulbs in the box.

There are {{{(matrix(2,1,12,2)) = 12!/(2!*(12-2)!) = (12*11)/(2*1) = 66}}} ways to choose 2 bulbs from the 12 in the box.

Thus, the probability of picking 2 defective bulbs is {{{10/66 = 5/33}}}

There are {{{(matrix(2,1,7,2)) = 7!/(2!*(7-2)!) = (7*6)/(2*1) = 21}}} ways to choose 2 non-defective bulbs from the seven non-defective bulbs in the box.

Thus, the probability of picking 2 non-defective bulbs is {{{21/66 = 7/22}}}

Finally, the probability of picking at least one defective bulb is

{{{1 - 7/22 = 15/22}}}