Question

. A consignment of 12 electronic components contains 1 component that is faulty. Two components are chosen randomly from this consignment for testing.

a. How many different combinations of 2 components could be chosen?

b. What is the probability that the faulty component will be chosen for testing?

Answer 1

a.

$$\dbinom{12}{2}=\dfrac{12!}{2!(12-2)!}=66$$

b.

$$\dfrac{\dbinom{1}{1}\dbinom{11}{1}}{\dbinom{12}{2}}=\dfrac{11}{66}=\dfrac{1}{6}$$