This note comes from Introduction to Probability, 2nd Edition
Example 1.9 Rada Detection
If an aircraft is present in a certain area, a radar detects it and generates an alarm signal with probability 0.99. If an aircraf is not present. the radar generates a (false) alarm, with probability 0.10. We assume that an aircraft is present with probability 0.05. What is the probability of no aircraf presence and a false alarm? What is the probability of aircraf presence and no detection?
$A$ sequential representation of the experiment is appropriate here, as shown in Fig. 1.9. Let $A$ and $B$ be the events $A = {an\ aircraft\ is\ present}$, $B = {the\ radar\ generates\ an\ alarm} $, and consider also their complements $A^c={an\ aircraft\ is\ not present}$$，$$B^c={the\ radar\ does\ not\ generate\ an\ alarm}$。
The given probabilities are recorded along the corresponding branches of the tree describing the sample space, as shown in Fig. 1.8. Each event of interest corresponds to a leaf of the tree and its probability is equal to the product of the probabilities associated with the branches in a path from the root to the corresponding leaf. The desired probabilities of false alarm and missed detection are
$$P(false\ alarm) = P(A^c ∩ B) = P(A^c)P(B | A^c) = 0.95 · 0.10 = 0.095$$，
$$P(missed\ detection) = P(A ∩ B^c) = P(A)P(B^c | A) = 0.05 · 0.01 = 0.0005$$.
Extending the preceding example, we have a general rule for calculating various probabilities in conjunction with a tree-based sequential description of an experiment. In particular:
- (a) We set up the tree so that an event of interest is associated with a leaf. We view the occurrence of the event as a sequence of steps, namely, the traversals of the branches along the path from the root to the leaf.
- (b) We record the conditional probabilities associated with the branches of the tree.
- (c) We obtain the probability of a leaf by multiplying the probabilities recorded along the corresponding path of the tree.
multiplication rule
In mathematical terms, we are dealing with an event A which occurs if and only if each one of several events $A_1, . . . , A_n$ has occurred, i.e., $A = A_1 ∩ A_2 ∩ · · · ∩ A_n$. The occurrence of $A$ is viewed as an occurrence of $A_1$, followed by the occurrence of $A_2$, then of $A_3$, etc, and it is visualized as a path on the tree with $n$ branches, corresponding to the events $A_1, . . . , A_n$. The probability of $A$ is given by the following rule (see also Fig. 1.9).
The multiplication rule can be verified by writing
$$P(\cap^n_{i=1} A_i)=P(A_1)\frac{P(A_1\cap A_2)}{P(A_1)}\frac{P(A_1\cap A_2\cap A_3)}{P(A_1\cap A_2)}\cdots\frac{P(\cap_{i=1}^n A_i)}{P(\cap^{n-1}_{i=1} A_i)}$$,
and by using the definition of conditional probability to rewrite the right-hand side above as
$$P(A_1)P(A_2|A_1)P(A_3|A_1\cap A_2)\cdots P(A_N|\cap^{n-1}_{i=1} A_i)$$.
The intersection event $A = A_1∩A_2∩· · ·∩A_n$ is associated with a path on the tree of a sequential description of the experiment. We associate the branches of this path with the events $A_1, . . . , A_n$, and we record next to the branches the corresponding conditional probabilities.
The final node of the path corresponds to the intersection event $A$, and its probability is obtained by multiplying the conditional probabilities recorded along the branches of the path
$$P(A_1\cap A_2\cap\cdots\cap A_3)=P(A_1)P(A_2|A_1)\cdots P(A_n|A_1\cap A_2\cdots \cap A_{n-1}).$$
Note that any intermediate node along the path also corresponds to some intersection event and its probability is obtained by multiplying the corresponding conditional probabilities up to that node. For example, the event $A_1 ∩ A_2 ∩ A_3$ corresponds to the node shown in the figure, and its probability is
$$P(A_1\cap A_2\cap A_3)=P(A_1)P(A_2|A_1)P(A_3|A_1\cap A_2).$$
For the case of just two events, A1 and A2, the multiplication rule is simply the definition of conditional probability.
