Random Incidence Paradox Question

179 Views Asked by At

10 friends have a collection of 1000 candies, 15 have collection of 1300 candies, and 5 have collection of 600 candies. What is the average number of candies your friends have? If you mix all these coins and randomly pick one out, what is the expected size of collection that candy's owner has.

For the first part of question, I calculated as : (10*1000 / 10) +(15*1300 / 15)+ (5*600 / 5) which turns out to be 1083. However, I am confused about second part of question. If total candies are 32500 and the probability that you pick it out from first part of friends : (10*1000 / 32500). And similarly for the rest, (15*1300 / 32500), (5*600 / 32500). Then you multiply probability with count in that group. Final equation becomes:

0.3*10*1000 + 0.6*15*1300 + 5*600*0.092 = 14976. But I am not really sure about my answer

1

There are 1 best solutions below

0
On BEST ANSWER

Your answer for the first part is correct. However, for the second part we want to know how many coins the owner has. Note the following:

10000/32500 candies belong to the owner who has 1000 candies

19500/32500 candies belong to the owner who has 1300 candies

3000/32500 candies belong to the owner who has 600 candies

The above values are all probabilities so just multiply them with the number of candies the owner has. For an intuition note that there is 10000/32500 chance that the candy you choose belongs to the owner who has 1000 candies. Just use this to calculate the expected value.