Hi I am trying to solve the following question but couldnt figure out if i am moving in the right direction or not. The question is as follows.
I have a large number of candies. One day i decided to distribute them to my friends. Suppose 10 of my friends have a collection of 1000 candies each, the other 15 have a collection of 1300 candies each while the rest of your 5 friends have collected 600 candies each.
What is the average number of candies your friends have?
If you mix all of these candies and randomly pick one out, what is the expected size of the collection that candies's owner has?
I am trying to approach this problem Through Random Incidence Paradox. Am i doing correct can any one help me out here.
Simply the weighted average: $\frac{10 \times 1000 + 15*1300 + 5*600}{10+15+5} = 1083.33$
Group 1 brings $10 \times 1000 = 10000$ candies to the mix, group 2 brings $15 \times 1300 = 19500$ candies to the mix, and group 3 brings $5 \times 600 = 3000$ candies to the mix. There are a total of 32500 candies. You will pick a candy from an friend from group 1 with probability $10000/32500 = 0.3076$, group 2 with probability 0.6, and group 3 with probability 0.0923. Thus, the owner will have a candy collection of 1000 w.p. 0.3076, 1300 w.p. 0.6 and 600 w.p. 0.0923. You can now compute the expected value as $0.3076 \times 1000 + 0.6 \times 1300 + .0923 \times 600 = 1143.08$, which is not the same as part 1.
The random incidence paradox is a good framework to understand the details of this problem.