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

Answer:

$\dfrac{2}{7}$

Step-by-step explanation:

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

A customer can select 2 plates in

$C^{15}_2=\dfrac{15!}{2!(15-2)!}=\dfrac{15!}{13!\cdot 2!}=\dfrac{13!\cdot 14\cdot 15}{2\cdot 13!}=7\cdot 15=105$

different ways.

2 plates can be selected from the same dinner set in

$3\cdot C^5_2=3\cdot \dfrac{5!}{2!(5-2)!}=3\cdot \dfrac{3!\cdot 4\cdot 5}{2\cdot 3!}=3\cdot 2\cdot 5=30$

different ways.

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

$Pr=\dfrac{30}{105}=\dfrac{6}{21}=\dfrac{2}{7}$