Chances of getting a present!

After understanding the classic break a piece of spaghetti into three pieces and tell me the probability that I can create a triangle problem I was inspired to create this problem. I think this problem captures the main gist in an easier to understand scenario.

Here is my post about the above problem: Spaghetti and Meatballs… I mean Triangles

Here’s the new problem:

I have wrapped 2 gifts for each of my 3 friends and have also wrapped some empty boxes for  each of them as well. In total I’ve wrapped 7 boxes for Alice, 5 boxes for Bob, and 8 boxes for Chris. If I choose one of my friends at random, and they randomly choose a present to open, what is the probability that there will be a gift inside?

I believe that understanding how this problem works and is solved will help in understanding why the Spaghetti Triangle problem is  solved the way it is. I’ve added some new comments on the bottom of my post on that problem as well.

I’ll add the solution later…

It’s later:

I choose Alice 1/3 of the time and 2/7 of the time Alice’s present will contain a gift. 1/3 times 2/7 is 2/21.

I choose Bob 1/3 of the time and 2/5 of the time Bob’s present will contain a gift. 1/3 times 2/5 is 2/15.

I choose Chris 1/3 of the time and 2/8 of the time Chris’s present will contain a gift. 1/3 times 2/8 is 2/24 = 1/12.

2/21 + 2/15 + 1/12 = .3119

The answer is not 6/20 = 3/10 = 30% because even though there were 6 gifts in 20 wrapped presents, once you chose a person the chances of getting a gift is different. 

It is the same with the spaghetti problem because each piece creates a different length to find the probabilities of a second break. This second break is the similar to a person having a different number of presents.