Law of total probability explanation What is the intuition behind the law of total probability?
http://en.m.wikipedia.org/wiki/Law_of_total_probability
 A: The idea is that if you divide the entire probability space into some pieces $B_n$, and then you consider another event $A$, you can also divide up $A$ into pieces based on how the event overlaps with the $B_n$. Perhaps a picture would help. The red lines split up the space of the black box into pieces. The blue oval stands for the event $A$. Notice how we can calculate the area of $A$ by adding up the areas of all the green pieces.

A: I would try to elaborate on Peter's answer, using Peter's example with the different pieces named:

Let $\Omega$ be the entire sample space, i.e. all of the possible outcomes of our experiment. In the image, this is the entire rectangle, while each possible outcome is a point in the rectangle.
Sticking to Peter's notation, we denote $B_{n}=\left\{ B_{1},...,B_{7}\right\}$.
For every outcome of the experiment, either $A$ occurs or doesn't occur. In the image, $A$ occurs for any outcome in a green area, and doesn't occur for any outcome in a white area.
The law of total probability gives us a way to calculate $\text{Pr}\left(A\right)$.
Here is one way to define $\text{Pr}\left(A\right)$:
$N$ scientists independently perform our experiment. 
Let $X$ be the number of scientists such that $A$ occurred for their outcome of the experiment.
We define: $$\text{Pr}\left(A\right)=\underset{N\rightarrow\infty}{\text{lim}}\frac{X}{N}$$
(This seems to make sense, as when we say that the probability of a fair coin to turn up tails is $0.5$, we mean that as the number of flips "approaches $\infty$", the proportion 
$\frac{\text{number of tails}}{\text{number of flips}}$ should be $0.5$)
For every $i\in\left\{ 1,...,7\right\}$, let $X_i$ be the number of scientists such that both $A$ and $B_i$ occurred for their outcome of the experiment.
$B_n$ are disjoint, so for every scientist that we count in $X$, we count that scientist at most in one of $\left\{ X_{1},...,X_{7}\right\}$.
Moreover, the union of $B_n$ is $\Omega$. Therefore, for every scientist that we count in $X$, we count that scientist at least in one of $\left\{ X_{1},...,X_{7}\right\}$.
Thus, for every scientist that we count in $X$, we count that scientist exactly in one of $\left\{ X_{1},...,X_{7}\right\}$.
Lastly, for every scientist that we don't count in $X$, we also don't count that scientist in any of $\left\{ X_{1},...,X_{7}\right\}$.
From the last two conclusions we can deduce that $X=X_{1}+...+X_{7}$.
Using the same definition that we used for $\text{Pr}\left(A\right)$, for every $i\in\left\{ 1,...,7\right\}$ we get:
$$\text{Pr}\left(A\cap B_{i}\right)=\underset{N\rightarrow\infty}{\text{lim}}\frac{X_{i}}{N}$$
Thus:
$$\begin{gathered}\text{Pr}\left(A\right)=\underset{N\rightarrow\infty}{\text{lim}}\frac{X}{N}=\underset{N\rightarrow\infty}{\text{lim}}\frac{X_{1}+...+X_{7}}{N}=\underset{N\rightarrow\infty}{\text{lim}}\frac{X_{1}}{N}+...+\underset{N\rightarrow\infty}{\text{lim}}\frac{X_{7}}{N}=\text{Pr}\left(A\cap B_{1}\right)+...+\text{Pr}\left(A\cap B_{7}\right)\\
\downarrow\\
\text{Pr}\left(A\right)=\text{Pr}\left(A\cap B_{1}\right)+...+\text{Pr}\left(A\cap B_{7}\right)
\end{gathered}
$$
Which is one version of the law of total probability (for our experiment).



The other version of the law (for our experiment) is:
$$\text{Pr}\left(A\right)=\text{Pr}\left(A|B_{1}\right)\text{Pr}\left(B_{1}\right)+...+\text{Pr}\left(A|B_{7}\right)\text{Pr}\left(B_{7}\right)$$
One can interpret this version as claiming that $\text{Pr}\left(A\right)$ is a weighted average of $\left\{ \text{Pr}\left(A|B_{1}\right),...,\text{Pr}\left(A|B_{7}\right)\right\}$, while for every $i\in\left\{ 1,...,7\right\}$, the weight of $\text{Pr}\left(A|B_{i}\right)$ is determined by $\text{Pr}\left(B_{i}\right)$.
Intuitively, this makes sense, as it means that more likely events are more heavily weighted.
By the way, a nice property that we can deduce from the weighted average interpretation is:
$$\text{min}\left\{ \text{Pr}\left(A|B_{1}\right),...,\text{Pr}\left(A|B_{7}\right)\right\} \le\text{Pr}\left(A\right)\le\text{max}\left\{ \text{Pr}\left(A|B_{1}\right),...,\text{Pr}\left(A|B_{7}\right)\right\}$$
