ANSWERS
a. 1/364
b. 165/364
Step-by-step explanation
a. There are 11 non-defective transistors and 3 defective transistors in the box. When the first one is selected, there are 3 defective transistors and 14 transistors in total. When the second one is selected, assuming that the first was defective, there are 2 defective and 13 transistors in total. Finally, if the first two were defective, when the third transistor is selected there will be only 1 defective transistor out of 12 transistors in total.
Therefore, the probability that the three transistors selected are defective is,
![P(all\text{ }defective)=(3)/(14)\cdot(2)/(13)\cdot(1)/(12)=(6)/(2184)=(1)/(364)]()
b. Now, we have to find the probability that the three transistors selected are non-defective.
First, we have to find how many ways are to select 3 transistors out of the 14 total transistors,

Also, we have to find how many ways are to select 3 non-defective transistors - note that these must be picked out of the 11 non-defective ones,

Therefore, the probability that the three transistors selected are non-defective is,
![P(none\text{ }defective)=(165)/(364)]()