Question 721547: Suppose that a restaurant includes a free toy with every kid’s meal. There are five types of toys, and a kid wants to collect at least one of each type; let X be the number of meals required for the kid to succeed. Assume that at each meal, all five types of toys are equally likely, independent of previous meals.
1.Find a natural way to express X as a sum of independent random variables.
Answer by Positive_EV(69)   (Show Source):
You can put this solution on YOUR website! For the first toy the kid gets, he has a 5/5 (1) chance of getting a new toy. The first term in your sum is 1.
For the second toy, there is a 4/5 chance he'll get a new toy. The number of toys required to get a second toy is geometrically distributed with p = .8. Note I am using the form of the geometric distribution where a success on trial 1 corresponds to x = 1, not x = 0.
The third toy, there's 3 unique toys left to get, so there is a 3/5 chance from each meal. The third toy's number of attempts is geometric(p = .6). Likewise, the 4th toy has a 2/5 chance of coming, so the number of tries is geometric(p = .4), and the last toy has a 1/5 chance, so it is geometric(p = .2).
So the number of trials required to get the toy can be expressed as the sum 1 + geometric(.8) + geometric(.6) + geometric(.4) + geometric(.2). As an aside, the mean number of meals the kid will need to buy is 11 5/12 meals.
|
|
|
