Carcass wrote:

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?

A. \(\frac{2}{7}\)

B. \(\frac{2}{5}\)

C. \(\frac{2}{3}\)

D. \(\frac{5}{6}\)

E. \(\frac{3}{2}\)

Solution:

Selecting 2 plates from all the plates ( 3 different sets and each having 5 plates = 3x5 = 15) => 15c2 = (15x14)/2 = 15 x 7

Selecting 2 plates from same set = Selecting 1 set from 3 sets (3c1) and Selected 2 plates from the set selected (5c2).

i.e; 3c1 x 5c2 = 30

Probability = (3c1 x 5c2) / 15c2

= 30/ (15 x 7)

= 2/7