I am studying Probability Theory 1, and we have learned the Law of total probability and proved it. I know the theoretical intuition behind it and I know why it makes sense from it's proof. But when I see a new problem (that they solved it using this method) I just can't relate it to this law.
Can you guide me of how and when should I use it, and what kind of problems can be solved using this theorem? I would like some simple example illustrating it's usage.
Suppose you do this experiment. Roll a fair pair of dice, and note the number that appears. Then, toss a fair coin that number of times. What is the probability that you get exactly one head? You need to break the outcomes down according to the number appearing on the pips. This is a good example of the use of the law of total probabilities. Often it is visualized using probability trees.
The basic idea behind the law of total probability is that if you have a family of disjoint events $\{A_i\}_{i=1}^n$ that cover the entire sample space, then for any event $B$ we have $$P(B)=P(B|A_i)P(A_i)+\cdots+ P(B|A_n)P(A_n) $$ or, if you don't like conditional probabilities $$P(B)=P(B \cap A_1)+ \cdots + P(B \cap A_n). $$
For instance, suppose a drawer contains one 4-sided, two 6-sided and one 8-sided dice. An unknown die is chosen at random, and you are told that it rolled a 3. What are the odds the die is 6-sided?
Here's a natural partition of the sample space of possible dice $$D_4=\text{'die is 4-sided'} \\ D_6=\text{'die is 6-sided'} \\ D_8=\text{'die is 8-sided'}$$
Let's define the additional event $$R_3=\text{'die rolled a 3'} $$ we are asked to find $$P(D_6|R_3).$$ It's typical to use Bayes' theorem in this case: $$P(D_6|R_3)=\frac{P(R_3|D_6) P(D_6)}{P(R_3)}.$$ One easily sees that $$P(R_3|D_6)=\frac{1}{6},\;P(D_6)=\frac{1}{2}$$ in order to find the denominator, we use the law of total probability: $$ \begin{align} P(R_3)&=P(R_3|D_4)P(D_4)+P(R_3|D_6)P(D_6)+P(R_3|D_8)P(D_8) \\ &=\frac{1}{4} \cdot \frac{1}{4}+\frac{1}{6} \cdot \frac{1}{2}+\frac{1}{8} \cdot \frac{1}{4}=\frac{17}{96}. \end{align}$$ All in all, we find $$ P(D_6|R_3)=\frac{8}{17},$$ which is smaller then the prior probability $P(D_6)=\frac{1}{2}$.
Remark: It is common to use the law of total probability in order the evaluate the denominator in Bayes' theorem.
The examples given by the other guys clearly illustrate the use of the law. To put it more informally (and perhaps more simply), just check if the main experiment in the problem naturally breaks down into a sequence of sub-experiments. Usually in the elementary exercises in probability texts, there would be 2 such sub-experiments, one followed by the other. You would have to compute the probabilities in each sub-experiment and then put them together using the Law of Total Probability to get the answer. The probability tree is a formally precise description that can be used for such a multi-stage experiment.
