Why is the denominator of Bayes theorem difficult to calculate on its own?

Question

According to wiki:
http://en.wikipedia.org/wiki/Normalizing_constant

The denominator of the Bayes theorem formula should be the probability of producing the data, but on its own is difficult to calculate. So how come it's difficult to calculate? And why can't we calculate it by finding the joint priori probability of the evidences?

Please refer to my another question:
http://answers.yahoo.com/question/index;_ylt=AmcLvBNJBUk31tqpMNlK1BDsy6IX;_ylv=3?qid=20070927031725AAAfbNB

I have been stuck in this denominator for a few weeks, please help.
Thank you very much. (don't give me advertisements)

I understand most of your calculation, but in the end, why is P(B) = P(B1) + P(B2) + P(B3)?
Thanks.

Answer 1

For the other question about Bayes formula

P(A|B) = P(A∩B) / P(B)

If B = B1 U B2 U B3 with B1∩B2 = Ø, B2∩B3 = Ø and B1∩B3 = Ø (isn't it ?)

then P(A∩B) = P(A∩B1) + P(A∩B2) + P(A∩B3)
=P(A|B1)*P(B1) + P(A|B2)*P(B2) + P(A|B3)*P(B3)

and P(A|B) = [P(A|B1)*P(B1) + P(A|B2)*P(B2) + P(A|B3)*P(B3)] / P(B)

In this case (only) P(B) = P(B1) + P(B2) + P(B3)

otherwise, what is the relation between B and B1, B2, B3 ?

Additional details :

P(CuD) = P(C) + P(B) - P(C∩D)
if C∩D = Ø
then P(C∩D) = 0 and P(CuD) = P(C) + P(D)

It is the same result with 3 events :
if B = B1 u B2 u B3 with B1∩B2 = Ø, B2∩B3 = Ø and B1∩B3 = Ø
then P(B) = P(B1) + P(B2) + P(B3)

for example : if we have an urn with 20 balls, 3 are red, 4 are blue, 5 are yellow and 8 are green.
We take one ball
The probality to have a ball no blue is :
P(red or yellow or green) = P(red) + P(yellow) + P(green) = 3/20 + 5/20 + 8/20 = 16/20
I can do so, because a ball can't be red and yellow, at the same time ...

In a group of 20 students, 4 are learning spanish, 5 french, 6 chinese and 7 german.
I can't use the same formula because a student can learn spanish and french, or 3 different languages ...

Answer 2

I will consult my decision tree to answer that.