Three light bulbs are chosen at random from $15$ bulbs, of which five are defective. What is the probability that:

1. None of the three are defective.

   Answer (a):

   $\frac{_{10}C_3 \cdot _5C_0}{_{15}C_3}=\frac{24}{91}$

2. Exactly two are defective.

   Hint (b):

   Exactly two are defective means that one is not defective.

   More Help (b):

   Use the method from part (a).