Total probability with different probabilities

Pengwuino
Gold Member
Messages
5,112
Reaction score
20
I have a silly question that has actually been kinda racking my brain since I have almost no formal education when it comes to probabilities. Let's say I know that a girl has a 40% chance of earning a college degree. Let's also say that right handed people have a 30% chance of earning a college degree. Then let's keep tossing in things like california girls have a 50% chance of getting a college degree. Obviously you can keep finding characteristics of that girl and find the statistics behind their success in getting a college degree.

Now the grand question is... what is the probability that a californian, right-handed, girl has of getting a college degree? I assume we can't simply multiply probabilities because it seems like you can keep throwing probabilities out like that and eventually make the probability nearly 0 even though it can't be...
 
Physics news on Phys.org
A = person goes to college, B = person is female, C = person is right-handed
You are given P(A|B) and P(A|C) and wish to find P(A|B,C). Let's assume that B,C are independent, and also assume that B,C are conditionally independent given A--that is, P(B,C|A) = P(B|A)P(C|A), and P(B,C) = P(B)P(C).
So,
P(A|B,C)=P(A,B,C)/P(B,C) = P(A)P(B,C|A)/P(B,C) = P(B|A)P(C|A) P(A)/(P(B)P(C)) = P(A|B)P(B)/P(A) P(A|C)P(C)/P(A) P(A)/(P(B)P(C)) = P(A|B)P(A|C)/P(A)

So instead of just multiplying by P(A|C), you multiply by P(A|C)/P(A)
 
mXSCNT said:
So instead of just multiplying by P(A|C), you multiply by P(A|C)/P(A)

Or, cutting out all the Baysian stuff (good, but if you haven't had a probability class I don't imagine it would make sense):

If a foo has a 30% chance of getting a degree, but a person (including foos and non-foos) has a 35% of getting a degree, you should be multiplying by .30/.35 rather than .30.
 
Yah this is what sucks about never having taken a statistics class :(.

So let me make the problem more accurate and tell me if i do this right. A girl has a 40% chance of having a bachelors degree. Girls who are right handed have a 50% chance of having a degree. Girls who live in california have a 30% chance of having a degree.

Thus the probability that a girl who is right handed who lives in california will have a (.4*.4*.4)/(.3*.5) chance of having a degree? Thus a 42.6% chance?

What if say, the 2nd condition "girls who are right handed have a 50% chance of having a degree" was changed to "people who are right handed have a 50% chance of having a degree"?
 
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Thread 'Detail of Diagonalization Lemma'
The following is more or less taken from page 6 of C. Smorynski's "Self-Reference and Modal Logic". (Springer, 1985) (I couldn't get raised brackets to indicate codification (Gödel numbering), so I use a box. The overline is assigning a name. The detail I would like clarification on is in the second step in the last line, where we have an m-overlined, and we substitute the expression for m. Are we saying that the name of a coded term is the same as the coded term? Thanks in advance.

Similar threads

Replies
7
Views
1K
Replies
11
Views
2K
Replies
1
Views
2K
Replies
11
Views
3K
Replies
3
Views
2K
Replies
41
Views
5K
Replies
8
Views
1K
Replies
1
Views
1K
Back
Top