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. \(\frac{3}{2}\)
Another way..
Any of the 15 can be picked up in the first go. so P = \(\frac{15}{15}=1\)
Next can be any of the remaining 10 to ensure they are from different sets so P = \(\frac{10}{14}\)
Over all prob = \(1*\frac{10}{14}=\frac{5}{7}\)
Thus required probability = 1- \frac{5}{7}=\frac{2}{7}[/m]