Question

A box at a yard sale contains 3 different china dinner sets, each consisting of 5 plates. A customer will randomly select 2 plates to check for defects. What is the probability that the 2 plates selected will be from the same dinner set?

Answer 1

`(2)/(7)`

Explanation:

3 different china dinner sets, each consisting of 5 plates consist of 15 plates.

A customer can select 2 plates in

`C^(15)_2=(15!)/(2!(15-2)!)=(15!)/(13!\cdot 2!)=(13!\cdot 14\cdot 15)/(2\cdot 13!)=7\cdot 15=105`

different ways.

2 plates can be selected from the same dinner set in

`3\cdot C^5_2=3\cdot (5!)/(2!(5-2)!)=3\cdot (3!\cdot 4\cdot 5)/(2\cdot 3!)=3\cdot 2\cdot 5=30`

different ways.

Thus, the probability that the 2 plates selected will be from the same dinner set is

`Pr=(30)/(105)=(6)/(21)=(2)/(7)`