+6 There are 6 balls numbered 1 through 6 in a box and I take out two different balls at random. What is the expected value of the product of the two numbers on the balls?
+195 There are total of 6×5=30 ways of choosing two distinct balls out of 6 balls. Out of these 30 outcomes, only following 10 outcomes result in a product greater than 1
Favorable OutcomesProduct of Numbers on Balls
(1, 2)2
(1, 3)3
(1, 4)4
(1, 5)5
(1, 6)6
(2, 3)6
(2, 4)8
(2, 5)10
(2, 6)12
(3, 4)12
The expected value is the average of the values of each outcome. In this case, this means the average of the products of the numbers on the balls you draw.
So, the expected value is $ \frac{2 + 3 + 4 + 5 + 6 + 6 + 8 + 10 + 12 + 12}{30} = 34/15.
