Expected value of a selection with displacement

Problem: A box contains a yellow ball, an orange ball, a green ball, and a blue ball. Billy randomly selects 4 balls from the box (with replacement). What is the expected value for the number of distinct colored balls Billy will select?

This is the answer given in the site where this question was originally asked

I can't make sense of the last step. Am I missing something obvious? Why does summation of the expectancy of every individual colored ball give us the expectancy of distinct colored balls that are picked?

Can someone explain the connection?

• Linearity of expectation is just a theorem — what are you confused about? Sep 17 '20 at 11:13
• Im trying to associate the physical meaning with the summation. Expectancy of one colored ball is roughly the number of balls of that color being picked. Now why does summing that up give us the number of distinct colored balls that will get picked? Sep 17 '20 at 11:27
• Because there's only one of each colour Sep 17 '20 at 12:25
• Thanks for responding. Turns out I misunderstood the summation process itself. Sep 17 '20 at 15:18

What happens here is the same thing. Each ball has a RV that indicates if it has been seen already. Each RV has their own expectation on $$n$$ trials. Summing all RV will give you a RV whose expectancy is the sum of the respective expectancies. The sum of the 4 RV that tell you whether the balls has been seen will be the RV that tell you all of them have been seen.