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I have difficult to understand the following law. Can anyone explain it to me using a simple example that can be understood by an 8th grade student.

Let $A_1, A_2, \ldots, \text{ and } A_n$ be mutually exclusive and exhaustive events. Then for any other event $B$

$$P(B)=P(A_1B_1) + P(A_2B_2) + \cdots + P(A_nB_n) \enspace. $$

I assume that $P(A_1B_1)$ is the same as $P(A_1 \cap B_1)$, right? Also, what are $B_1$, $B_2$, $\ldots$, $B_n$? I read information on law of total probability in Wikipedia. It is too hard for me to understand.

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    Are you sure that both $A$ and $B$ are indexed in the equation. It would seem to me that the equation should be $$P(B) = P(A_1 B)+P(A_2 B) + P(A_3 B).$$ Also, you are correct about $$P(A_1 B_1) = P(A_1 \cap B_1),$$ it's just quicker and neater to omit the intersection sign. If what I've written is correct, I can answer your question, otherwise, please elaborate on the context. – 2017-02-28
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    @David， both A and B are indexed in the equation, but they could be wrong because I noticed some errors in the book. – 2017-02-28

2 Answers 2

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Law of total probability easily appeals to intuition and not difficult to understand. Let $A$ be the statement (an event) that some email I receive tomorrow will invite me for a party. Let $B_1, B_2, B_3$ denote the events that an email I received was sent (1) before breakfast (2) between breakfast and lunch (3) after lunch.

Now as the 3 cases of time durations are exclusive and cover all possibilities.

So the prob of email invitation for a party is the sum of probability of email invitation sent before breakfast plus probability of email invitation sent between breakfast and lunch plus probability of email invitation sent after lunch.

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My guess is the book is wrong. We need for the sets that we break $B$ up into to "partition" the original set $B$. That is, we need to be sure that when we split $B$ we have sets $\{B_i\}_{i=1}^n$ such that $$B = \bigcup_{i=1}^n B_i \text{ and } B_i \cap B_j = \emptyset, i \neq j.$$ The reasons we do that are so that we can split up the probability of $B$ into simpler pieces. By the equality in the sets, we have that $$P(B) = P\left(\bigcup_iB_i\right)$$ and by the mutual exclusivity, we can break up the union into a summation, i.e., $$P\left(\bigcup_iB_i\right) = \sum_iP(B_i)$$ (see https://en.wikipedia.org/wiki/Mutual_exclusivity for details). Now, this works for any sets $B_i$ that partition $B$. In particular, we can consider sets which partition our ENTIRE sample space, let's call our sample space $S$ and this partition of $S$ would be $\{A_i\}_{i=1}^n$ where $$S = \bigcup_iA_i.$$ I'm not sure how familiar you are with basic set theory, but we can now consider the intersections $B\cap A_i.$ Since, for any element in $B$, we know that it must fall in some set $A_i$ (because these cover the entire sample space), we have that $$B = \bigcup_iBA_i.$$ Therefore, we can compute the probability of $B$ using the identity I listed above. Namely, $$P(B) = P\left(\bigcup_iBA_i\right) = \sum_iP(BA_i).$$

So what is a practical example of this. Let's say that we wanted to know the probability that people are unhealthy, call this $P(B)$, but we only had a survey to determine this probability from. In particular, the survey asks: (1), do you see a doctor, and (2) are you healthy. The results are given that 1000 people were surveyed and 300 people answered that they see a doctor and are unhealthy and 100 people answered that they do not see a doctor and are unhealthy. Now, because we have that everybody either sees a doctor or does not see a doctor, these sets cover the entire sample space, i.e., if $A_1 := \text{sees doctor}$ and $A_2 := \text{doesn't see doctor}$ then $S = A_1 \cup A_2.$ Additionally, a person can't see a doctor AND not see a doctor, i.e., $A_1 \cap A_2 = \emptyset$. So we have that these two sets partition the entire sample space. So, in order to find $P(B)$ we can instead consider $$P(B) = P(BA_1)+P(BA_2) = \frac{300}{1000} + \frac{100}{1000} = \frac{2}{5}.$$