my problem is the following:
Lets imagine we have a computer with 3 memories (m1, m2, m3).
When data is needed it is searched if m1, if not found in m1, it is searched in m2 and so on.
P(finding data on m1) = P(m1) is unknown, P(m2) = 0.6
Now the question is: What is P(m1) knowing that 76% of the time we do not need to go to m3, we found it in m2.
How can you apply Bayes law here?
If you do P(m1 | m1 v m2), which i believe is, P of finding it in m1 knowing that is either in m1 or in m2,
P(m1 | m1 v m2) = P(m2 v m1 | m1) * P(m2)/P(m1) = 0.76
but then P(m2 v m1 | m1) is 1 since being in m2 or m1 knowing that it is in m1 is trivial, and you get that P(m2)/P(m1) = 0.76 wich means that P(m1) = P(m2)*0.76 = 0.6*0.76 but that is not the correct answer...
What am I doing wrong?
thanks a lot for your time and sorry for my bad explanations, english not my mother tongue :p
Bayes' law is not needed.
Let $M_i = \text{" the data is in memory } m_i"$. We're given that $$\Bbb{P}(M_2) = 0.6$$ and, since the data is not in $m_3$ if and only if it is in $m1$ or $m2$, we also have the given value $$\Bbb{P}(\overline{M_3}) = \Bbb{P}(M_1 \cup M_2) = 0.76$$ Note also that since the data is in only one of the three memories, $$\Bbb{P}(M_1 \cap M_2) = 0$$
Now by the principle of inclusion-exclusion, $$\Bbb{P}(M_1 \cup M_2) = \Bbb{P}(M_1) + \Bbb{P}(M_2) - \Bbb{P}(M_1 \cap M_2) $$ which we can now solve for $\Bbb{P}(M_1)$:
$$\Bbb{P}(M_1) = 0.76 - 0.6 = 0.16$$