Estimating probabilities using Bayes rule? I am working on a past exam paper. I am given a data set as follows:
Hair {brown, red} = {B,R}, Height {tall, short} = {T,S} and Country {UK, Italy} = {U,I}
(B,T,U) (B,T,U) (B,T,I)
(R,T,U) (R,T,U) (B,T,I)
(R,T,U) {R,T,U) (B,T,I)
(R,S,U) (R,S,U) (R,S,I)
Question: Estimate the probabilities P(B,T|U), P(B|U), P(T|U), P(U) and P(I)
As the question states estimate, I am guessing that I don't need to calculate any values. Is it just a case of adding up how many times P(B,T|U) occurs over the whole data set e.g. (2/12) = 16%.
Then would the probability of P(U) be 0?
 A: To simplify let $A$ be the event that an individual has brown hair and is tall, and $B$ be the event that an individual is from the UK. We know from the Kolmogorov definition that the conditional probability of A given B is $P(A|B)=P(A \& B)/P(B)$. The probability that an individual has brown hair and is tall and is from the UK, $P(A \& B)$, is the proportion of individuals with these characteristics which is $2/12=0.16$. The probability of an individual being from the UK, $P(B)$, is the proportion of individuals from the UK which is $8/12=0.666$. Hence the probability of $A$ given $B$, $P(A|B)$, is $P(A \& B)/P(B)=0.16/0.666=0.25$. Therefore $P(B,T|U)=0.25$.
Finding $P(B|U)$ uses the same logic as above. $P(B \& U)=2/12=0.16$. $P(U)$ is the proportion of individuals from the UK which is $8/12=0.666$. Therefore $P(B|U)=0.25$.
$P(T|U)$ is the same logic again. $P(T \& U)=6/12=0.5$. $P(U)=8/12=0.666$. $P(T|U)=0.5/0.666=0.75$.
$P(U)$ and $P(I)$ are the proportion of individuals from the UK and Italy respectively. $P(U)=8/12=0.666$ and $P(I)=4/12=0.333$
A: P(B,T|U) means given that the country is the UK, what is the probability of both Brown Hair and Tall.  You need to sum all the people that are (B,T) in the UK, and then divide by the number of people in the UK.  P(U) is the probability that someone is in the UK. 
