### Home > A2C > Chapter 13 > Lesson 13.1.2 > Problem13-38

13-38.

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.

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

2. Exactly two are defective.

Exactly two are defective means that one is not defective.

Use the method from part (a).