Answer:
[tex]\dfrac{2}{7}[/tex]
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
[tex]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[/tex]
different ways.
2 plates can be selected from the same dinner set in
[tex]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[/tex]
different ways.
Thus, the probability that the 2 plates selected will be from the same dinner set is
[tex]Pr=\dfrac{30}{105}=\dfrac{6}{21}=\dfrac{2}{7}[/tex]
