A fête stall's wheel is divided into three sectors — red 120°, blue 90° and green 150°. The wheel is spun twice. What is the probability that it stops on the same colour both times?

The sectors are unequal, so each colour's probability is its angle out of 360°: red 1/3, blue 1/4, green 5/12, which correctly add to 1. The spins are independent, so P(same colour) = $(1/3{)}^2+(1/4{)}^2+(5/12{)}^2=16/144+9/144+25/144=50/144=25/72$. Treating the three sectors as equally likely gives $3\times (1/3{)}^2=1/3$, which would need every angle to be 120°. The value 47/72 is the complement, two different colours, and 25/144 keeps only the green-then-green case.