Clarifying probability definition

Question

I was reading in the book "What is Mathematics" by Courant/Robbins about the algebra of sets and their application to the theory of probability.

They give an example:

The three digits 1, 2, 3 are written down in random order. What is the probability that at least one digit will occupy its proper place.

So we wish to calculate $p(A + B + C)$ (OR).

They give then a formula for calulating:

$$p(A + B + C) = p(A) + p(B) + p(C) - p(AB) - p(AC) - p(BC) + p(ABC)$$

1) I find this confusing, because in another book I found this definition:

$$p(\text{A or B or C}) = p(A) + p(B) + p(C)$$

What's the reason here for two different definitions?

2) In the "What is Mathematics" book they proceed to calculate $p(A) = p(B) = p(C) = \frac{1}{3}$ which is clear to me.

But then they calculate $p(AB) = p(AC) = p(BC) = \frac{1}{6}$ (AND) which I dont understand how they found it. I wrote down the combinations, and of course the answer is correct, put I thought for example $p(AB) = \frac{1}{3} \times \frac{1}{3}$ (although that would give a nonsensical answer.)

So how is $p(AB) = p(AC) = p(BC) = \frac{1}{6}$ found mathemtically?

Answer 1

Consider the diagram below:

We wish to find $\Pr(A \cup B \cup C)$, the probability that an element is contained in set $A$ or set $B$ or set $C$.

If we simply add the probabilities $\Pr(A)$, $\Pr(B)$, and $\Pr(C)$, we add the probabilities $\Pr(A \cap B)$, $\Pr(A \cap C)$, and $\Pr(B \cap C)$ twice, once for each set in which the intersection of two sets is contained. We only want to add them once, so we must subtract $\Pr(A \cap B)$, $\Pr(A \cap C)$, and $\Pr(B \cap C)$ from $\Pr(A) + \Pr(B) + \Pr(C)$. However, if we do so, we will not have added $\Pr(A \cap B \cap C)$ at all. This is because we add it three times in the sum $\Pr(A) + \Pr(B) + \Pr(C)$, once for each set in which $A \cap B \cap C$ is contained, then subtract it three times when we subtract $\Pr(A \cap B)$, $\Pr(A \cap C)$, and $\Pr(B \cap C)$ from the total, once for each pair of sets in which $A \cap B \cap C$ is contained. Thus, we must add $\Pr(A \cap B \cap C)$ to the total. Therefore, $$\Pr(A \cup B \cup C) = \Pr(A) + \Pr(B) + \Pr(C) - \Pr(A \cap B) - \Pr(A \cap C) - \Pr(B \cap C) + \Pr(A \cap B \cap C)$$ In the special case that sets $A$, $B$, and $C$ are mutually disjoint,
$$P(A \cap B) = P(A \cap C) = P(B \cap C) = P(A \cap B \cap C) = 0$$ so $$\Pr(A \cup B \cup C) = \Pr(A) + \Pr(B) + \Pr(C)$$

As for the problem in Courant and Robbins' What is Mathematics?, notice that if any two of the three numbers $1, 2, 3$ are in their proper places, so is the third. Since there are $3! = 6$ possible permutations and only one of them has two numbers in their proper places, $$P(A \cap B) = P(A \cap C) = P(B \cap C) = \frac{1}{6}$$ Moreover, $$P(A \cap B \cap C) = \frac{1}{6}$$ since only one of the six permutations has all three numbers in their proper places.

Consider the list of the six permutations of $1, 2, 3$.

$1, 2, 3$

$1, \color{red}{3}, \color{red}{2}$

$\color{red}{2}, \color{red}{1}, 3$

$\color{red}{2}, \color{red}{3}, \color{red}{1}$

$\color{red}{3}, \color{red}{1}, \color{red}{2}$

$\color{red}{3}, 2, \color{red}{1}$

Numbers marked in red are not in their proper places. Two of the six permutations of $1, 2, 3$ have no number in their proper place (these are called derangements). The remaining four permutations have at least one number in its proper place, so they belong to $A \cup B \cup C$. Hence, $$\Pr(A \cup B \cup C) = \frac{4}{6} = \frac{2}{3}$$ We can also compute $\Pr(A \cup B \cup C)$ using our formula. Since $\Pr(A) = \Pr(B) = \Pr(C) = \frac{1}{3}$, $\Pr(A \cap B) = \Pr(A \cap C) = \Pr(B \cap C) = \Pr(A \cap B \cap C) = \frac{1}{6}$, we obtain \begin{align*} \Pr(A \cup B \cup C) & = \Pr(A) + \Pr(B) + \Pr(C) - \Pr(A \cap B) - \Pr(A \cap C) - \Pr(B \cap C) + \Pr(A \cap B \cap C)\\ & = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} - \frac{1}{6} - \frac{1}{6} - \frac{1}{6} + \frac{1}{6}\\ & = \frac{4}{6}\\ & = \frac{2}{3} \end{align*}

Answer 2

If by $A+B$ you mean the union $A\cup B$ and by $AB$ the intersection $A\cap B$, then the second "definition" (more of a theorem, really) applies when $A,B$ and $C$ are disjoint sets/events. In this case, $AB, AC, BC,$ and $ABC$ are all the empty set/event, so they have probability zero. The first definition is the correct definition and applies to all possible $A,B$ and $C$.

As for question 2, one can see that eg. $P(AB)=1/6$ by noticing that if $1$ and $2$ are placed in their right positions, then also $3$ is placed in its right position. So the configuration $AB$ is precisely one out of the total six configurations possible when permuting $1,2,$ and $3$.

Answer 3

Consider the simpler example:

The two digits 1, 2 are written down in random order. What is the probability that at least one digit will occupy its proper place.

Let's find it in two ways: frequency and formula.

$1) \ \mathbf{Frequency}$. There are $2!=2$ permutations: $$1,2 \ \ (\text{success}) \\ 2,1 \ \ (\text{failure}) $$ So, the probability of success is: $$P(1,2)=P(1\cap 2)=\frac{n(\text{success})}{n(\text{total})}=\frac12.$$ $2) \ \mathbf{Formula}$. Let $A=\{1 \ \text{occupies its proper place}\},B=\{2 \ \text{occupies its proper place}\}$. We want to find $P(A\cup B)$, which implies $1$ occupies its proper place or $2$ occupies its proper place or both occupy their proper places. The formula of addition for any two events is: $$P(A\cup B)=P(A)+P(B)-P(A\cap B)=\frac12+\frac12-\frac12=\frac12.$$ Note that $P(A)=P(A\cap B)=P(B)=\frac12$, because if $A$ happens, so does $B$.

Note that the formula of addition is based on the inclusion-exclusion principle of counting (see: https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle).

Now, in case of three numbers $1,2,3$, the formula of addition for any three events $A,B,C$ is: $$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C).$$