% of % - Please Help Me Prove My Friend Wrong Here is the situation:  My friend and I are at an impasse.  I believe I'm correct, but he's so damn stubborn he won't believe me.  Also, I'm not the most articulate at explaining things.  Hopefully some of you guys can help me explain this to him in a way he'll understand.  Here is the problem:
A DVD is either at his parent's house or his own.  The probability that it's at his house is 30%.
If the DVD is at his own house, there is a 90% chance it's on the porch, and a 10% chance it's in the living room.
What is the % chance the DVD is on the porch?
My friend says you take 90% of 30% which is 27 and that is the % chance it's on the porch.  Is this correct?  I don't believe so.  
I believe that regardless of where the DVD is, the chance of it being anywhere in his house is still 30% overall.  Location inside his house won't change those odd because the porch and the living room are both part of the house.  If there is a 90% chance it's on the porch, it doesn't change the overall odds of it being in that location.
Now, if you rephrase the question and ask, "The DVD Is either at my parents, my porch, or my living room.  What is the % chance it's on my porch?", the answer is 33%.  If there are three places it could be, then there is 33.333% chance it's on the porch.  Even if it's a 90% chance it's at his house, if there are only three places it can be, it remains the same.
I think the correct way of answering the question is:  There is a 30% chance the DVD is at my house.  If it is at my house, there is a 90% chance it's on my porch.  They are two separate odds and you can't take a percentage of the overall odds since the locations are inside the house. 
Is this correct or am I wrong?  And regardless, please give me your explanation.
 A: Your friend is correct and I'll give you an experiment you can try:
Take two boxes labeled "my house" and "my friend's house" and in the first box put two bags labeled "porch" and "living room". You will also need a marble and a die. Take one die and roll it (we're going to approximate 70% ~ 2/3 here). If it's a 1,2,3 or 4, put your hand in the "my house" box but don't pick a bag yet. If it's a 5 or 6 put it anywhere in the "friend's house" box. If your hand is in the "my house" box then you need to roll again to figure out which bag to put it in. If it's a 1,2,3,4 or 5 (let's say 90% ~ 5/6) put it in the "porch" box, and if it's a 6 put it in the "living room" box. Repeat this exercise until you have a feel for how often it lands in "my porch." Record the trials and see what the odds are. You should be convinced now.
Your reasoning is incorrect here:

If there is a 90% chance it's on the porch, it doesn't change the overall odds of it being in that location.

No, you changed things. There is not a 90% chance it's on the porch. If - if! - it is in your house, then (and only then!) there is a 90% chance it's on the porch. The magic of the 27% comes from the fact that mathematically - and experimentally, as I hope the above box/bag/marble exercise shows - we know how the "in your house" 30% and the "on the porch" 90% interact. Namely, they interact multiplicatively.
How about this? There is a 30% chance you'll go to New York and a 100% chance you'll go to the Empire State Building if you go to New York (because why else would you go to New York? kidding...). Does this mean there's a 100% chance you'll go to the Empire State Building? Well, since you have to go to NY to go to the ESB, that would mean there's a 100% you'll go to New York - and now we're being contradictory! So this interpretation makes no sense and is never what we mean mathematically or in plain English.

"The DVD Is either at my parents, my porch, or my living room. What is the % chance it's on my porch?", the answer is 33%.

This is definitely wrong. I'm pretty sure you either have ebola or you don't have ebola. Up to you whether you need to call 911 and get yourself quarantined right away because there's a 50% chance you have ebola. Or perhaps the doctors gave your sick relative 6 months to live, but your relative might live a year or two years or three years or four years or five years, which means there's at least an 86% chance the doctor is wrong.
Now put 10 red m&ms in a bag and 1 blue one. Grab one without looking. Since there's two possibilities, there's a 50% chance it's a blue one, right? So I'll bet you a dollar that it's red and you bet me a dollar that it's blue, and we'll see who's paying for lunch later.
A: 
I believe that regardless of where the DVD is, the chance of it being anywhere in his house is still 30% overall.

And it is! And 90% of the time that it is, it is on the porch, and 10% of the time it's in the living room.
That makes, in total, 90% of 30% is 27% chance that it's on the porch, and, a 10% of 30% is 3% chance that it's in the living room.
Since 27% + 3% = 30%, that means that the chance that it is anywhere in his house is indeed still 30% overall. That part of your intuition was fine.
It's just that that 30% can be subdivided in portions based on the probability of each place in the house it could be, if it is in the house.
A: The DVD is


*

*at his parents' house with a probability of $70\,\%$ ($=100\,\%-30\,\%$)

*in the porch of his house with $27\,\%$ ($=90\,\%$ of $30\,\%$)

*in the living room of his house with $3\,\%$ ($=10\,\%$ of $30\,\%$)


Check: $70\,\% + 27\,\%+3\,\%=100\,\%$.
A: Your friend is right, and his explanation is right.  Ask yourself this, using the reasning you put forth in your explanation:
My hat is either in my closet, on my head, or I left it on Mars.  What is the probability that it is on Mars?  
I'm guessing you can see that just because there are 3 possibilities, that does not mean we should assign a 33.33% probability to each.   
A: See, it's true that there are only three possibilities: at your house, the porch or the living room, but it is not equally likely that one may occur. Consider this: if we have to compute the probability that the today is Sunday, you might say 1/2, either it is a Sunday or it's not. But being both is not equally likely. The number of possibilities that it is not a Sunday is 6, and that it is a Sunday is 1. So the probability that today is a Sunday is 1/7.
Similarly the above goes in your case as well. There are two possibilities: either the DVD is in your house, or it is in your friends house. If it is in your friends house, it's either in the living room or the porch. The probability that it is in your friends house is $30%$, so the sub cases that exist here will have their sum as $30%$. The sub cases are $90%$ and $10%$ are percentages of the percentage of the actual value, which is $30%$. So it's 27 and 3 respectively.
A: You wanted confirmation of your view... well tough luck: You are confused and your friend is right.
I think this is as much a language as a math problem, the "%" ist confusing you, so do away with it for a moment.
To do this you need to do some/a lot running around (at least virtually/in your head):
Assume your friend and you are going to a pub every weekend, afterwards you watch a DVD together. You do this 100 times. 70 times, you need to go to his parents to get the DVD. 30 times you need to go to you friend's place to get it.
Now to get back to this "%"-thing... "percent" or (remember: language problem) perhaps think about it as "per cent", which means "per a hundred" or "from a hundred".
So 30 times from 100 times is 30%, right?
In 10% of the times you go to his place, the DVD is in the living room, and 10% from 30 is 3 times. Three times from a hundred, per a hundred, per cent, or 3%.
So the probabilites for your given problem are:


*

*70% DVD is at friend's parent's 

*27% DVD is on friend's porch 

*3% DVD is in friend's living room


To answer your general question, it's valid to calculate percent of percent to reach the overall probability.
A: Probabilities like this can be represented as a tree:
100% --- 70% Parent's House - 70%
      |
      |- 30% His House ------ 27% (30%*90%) Porch 
                           |-  3% (30%*10%) Living Room

Notice all the leaves of the tree add up to 100%.
The probability of it being on the porch is conditional on it being in his house.  Conditional probabilities are calculated based on the condition of the parent probability being true.  

They are two separate odds and you can't take a percentage of the
  overall odds since the locations are inside the house.

They are not separate.  There are four places described: parent's house, his house, porch, living room.  Consider these two statements, one is obviously wrong:


*

*The DVD can be at his house, and also be on the porch. (This makes sense because the porch is at his house.)

*The DVD can be at the parent's house, and also be on the porch. (Makes no sense, because the porch is not at his parent's house)


If they were truly separate as you rationalize, then either statement could be true.  However, only one of the above is true because there the chance of it being on the porch is conditional on it being in the house, and therefore is conditional on the probability of it being in the house.  So the probability of it being on the porch is a portion of the probability of it being in his house.
If we added another probability, such as "There is a 40% chance of the DVD being scratched." then that would be independent of where it was.  Thus it would be considered an independent probability and how you calculated the probability of combinations of where it was and whether or not it was also scratched would be quite different.
If you Google conditional probability, you will find alot of examples.
A: There are already really nice answers to your question that explain why you are wrong.
I am going to try and show you where your thinking goes wrong.
Lets say you have 2 stones. One is blue and the other is red.
Case 1: If I ask you to pick a stone, what is the probability that you will pick the red one?

 50%. since you have two options and you can only pick one. This gives you 1/2 aka 50%.

Now lets say I hide them. One in a black box and the other in a white box.
Case 2: And now I ask you to pick a box. What is the probability that you will pick a box with the red stone?

 Again 50%. 

Here is a list of all the possibilities for Case 2:

 1. White Box, Red stone
 2. White Box, Blue stone
 3. Black Box, Red Stone
 4. Black Box, Blue Stone

And since there are:

 2 out of 4 ways to pick a red stone, you have a 50% chance

Case 3: And what is the probability that you will find the red stone in the white box?

 25%! The difference here is that first you will have to pick a box and then you have to deal with the probability of finding the red rock in that box. 

You can list out all the possibilities for Case 3:

 1. White Box, Red stone
 2. White Box, Blue stone
 3. Black Box, Red Stone
 4. Black Box, Blue Stone

and you will find that only:

 1 out of the 4 possibilities gives you the correct combination. Hence 25%! This is the same as 50% to pick a box and 50% that you get the red rock.

What you and your friend are arguing about is similar to Case 3 but you are mixing it up with Case 2 and hence your friend is correct.
A: Your friend is right. Let $A$ = "at his house" and $B$ = "on his porch". Then $A \cap B$ = "at his house AND on his porch". $P\left(X\right)$ means "the probability that $X$ is true"; $P\left(X|Y\right)$ means "the probability that if $Y$ is true, then $X$ is also true". $P\left(X|Y\right)$ is referred to as a conditional probability.
You have given that $P\left(A\right) = .30$ and $P\left(B|A\right) = .90$
The very definition of conditional probability is that $P\left(B|A\right) = \frac{P\left(B \cap A\right)}{P\left(A\right)}$
And since it cannot both be on his porch and not be at his house, $B \cap A = B$
Therefore, $P\left(B|A\right) = \frac{P\left(B\right)}{P\left(A\right)}$; $P\left(B\right) = P\left(B|A\right)P\left(A\right)=\left(.90\right)\left(.30\right)=.27$
Think of it this way: X = "on his porch"; Y = "at his house but not on his porch"; Z = "not at his house" (Therefore, "either X or Y" = "at his house")
XXXXXXXXXX
XXXXXXXXXX
XXXXXXXYYY
ZZZZZZZZZZ
ZZZZZZZZZZ
ZZZZZZZZZZ
ZZZZZZZZZZ
ZZZZZZZZZZ
ZZZZZZZZZZ
ZZZZZZZZZZ

If you choose a random letter, then your random choice is consistent with the information you provided:


*

*there is a 30% chance that you will get either X or Y

*There is a 90% chance that if you get X or Y, then it is X
Note that 27 of the 100 letters are X.
A: You've conflated unconditional probability with conditional probability. 
You've said that 


*

*p(parents house) = 70% 

*p(friend's house) = 30%.   


Here p stands for probability. So far so good. Then you gave the conditional probabilities:


*p(porch | at friends house) = 90%

*p(living room | at friends house) = 10%


Here the vertical bars are common notation indicating that the probability that it is on the porch or living room is conditioned on it being at your friend's house.
In general the joint probability, p(A,B), is given by p(A,B) = p(A|B)p(B). In this case A is "porch" and B is "at friend's house":
p(porch, at friend's house) = p(porch|at friends house) x p(friends house) = .9 x .3 = 27%
A: Here's the minimal answer, with no philosophy or extraneous material.
Define the following events:


*

*its at your own house $=O$

*its on the porch (of your own house) $=O_p$
We're trying to find $\mathbf{P}(O_p).$
The following information is given.


*

*(Implicitly: $O_p \subseteq O$)

*$\mathbf{P}(O) = 0.3$

*$\mathbf{P}(O_p \mid O) = 0.9$


By the definition of conditional probability: $$\mathbf{P}(O_p \mid O) = \frac{\mathbf{P}(O_p \cap O)}{\mathbf{P}(O)}.$$
But since $O_p \subseteq O$, hence:
$$\mathbf{P}(O_p \mid O) = \frac{\mathbf{P}(O_p)}{\mathbf{P}(O)}.$$
Rearranging: $$\mathbf{P}(O_p) = \mathbf{P}(O_p\mid O)\mathbf{P}(O) = 0.9 \times 0.3 = 0.27$$
So the probability that its on the porch of your own house is $0.27$ (your friend was right).
A: If you have already dictated the odds, you can't then change them and treat the house and places as isolated cases. Your friend is right.
