jmc

Question

I have 5 marbles numbered 1 through 5 in a bag. Suppose I take out two different marbles at random. What is the expected value of the sum of the numbers on the marbles?

Accepted Answer

There are 52 = 10 different pairs of marbles can be drawn, and the expected value of the sum is the average of the sums of each pair. This is align* 110((1+2)+(1+3)+(1+4)+(1+5)+(2+3)& +(2+4)+(2+5)+(3+4)+(3+5)+(4+5))&=6010 = 6. align*