Probability axiom

The event $A$ and the event $B$ are the same. Then we write，$A = B$．The event $A$ is included in the event $B$. Then we write $A \subset B$．

• All events it is expressed by the sample space $\Omega$
• Sum event The event $A$ or the event $B$ can occur and denote by．$A \cup B$.
• Product events The event $A$ and $B$ can occur and denoted by $A \cap B$．
• Complementary event Event $A$ does not occur for $A$ and denoted by $\overline{A}$．
• Empty event Events that never happen and denoted by $\phi$．

For the operation of events, the following relational expression holds as in the case of sets.

1. Let $\Omega$ be all events and $\phi$ be empty event. Then
   $$\Omega = \Omega \cup \phi, \phi = \Omega \cap \phi$$

2. For any event $A$,
   $$A \cap {\overline A} = \phi, \ A \cup {\overline A} = \Omega$$

3. For any event $A,B,C$て
   

   $$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$$

   $$A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$$

   
   
4. For any event $A,B$（DeMorgan's law)
   $$(\overline {A \cup B}) = {\overline A} \cap {\overline B}, \ \overline {(A \cap B)} = {\overline A} \cup {\overline B}$$

5. For any events $A$，$B$，$C$，
   $$(A \cap B) \cap (A \cap {\overline B}) = \phi, \ A = (A \cap B) \cup (A \cup {\overline B})$$

Theorem 2..1 (Probability axiom)   When the real number $P(A)$ that satisfies the following axiom is determined for any event $A$ in the sample space $\Omega$, $P(A)$ is called the probability of the event $A$. An event with a possible probability is called a probability event.

1. $$0 \leq P(A) \leq 1$$

2. $$P(\Omega) = 1, P(\phi) = 0$$

3. For a sequences of events $A_{1},A_{2},\ldots,A_{i} \ldots$, if any two of them are mutually exclusive
   $$P(\cup_{i=1}^{\infty} A_{i} = \sum_{i=1}^{\infty}P(A_{i})$$
   This property is called bf full additivity for probability.

Fix event $A$ and as a function of event $B$, we define

$$P(B\vert A) = \frac{P(A \cap B)}{P(A)}$$

Then this function satisfies the probability axiom． This is called the bf conditional probability of the event $B$ when the event $A$ occurs

If the events $A_{1},A_{2},\ldots,A_{n}$ are mutually exclusive and

$$A_{1} \cup A_{2} \cup \cdots \cup A_{n} = \Omega$$

Then for any event $B$,

$$B = B \cap \Omega = (B \cap A_{1}) \cup (B \cap A_{2} \cup \cdots \cup (B \cap A_{n})$$

Thus,

$$P(B) = P(B \cap A_{1}) + P(B \cap A_{2}) + \cdots + P(B \cap A_{n})$$

Note that by the conditional probability, $P(B \cap A_{i}) = P(A_{i})P(B \vert A_{i})$. Thus we can find

$$P(B) = P(A_{1})P(B \vert A_{1}) + P(A_{2})P(B \vert A_{2}) + \cdots P(A_{n})P(B \vert A_{n})$$

This is called Bayes's theorem．

Exercise 3

1. Let $A$ = [Throw the dice 4 times and get a 6 at least once]. $B$ = [Throw two dice 24 times at the same time and get a 6 at least once].

(a)

Find $P(A)$．

(b)

Find $P(B)$．

2. A patient has complained of certain symptoms. From the experience of doctors, we know that about $5\%$ of people in the same age group have cancer when they complain of the condition. On the other hand, a detailed examination shows a positive reaction of $85\%$ for true cancer patients and a positive reaction of $5\%$ for non-cancer patients. If a patient gives a positive result on the work-up, find the probability that the patient has cancer.．

3. Show the following relations.．

(a)

$P(\overline{A \cup B}) = P(\overline{A} \cap \overline{B})$

(b)

$P(\overline{A \cap B}) = P(\overline{A} \cup \overline{B})$

(c)

For $B$ and $C$ are mutually exclusive， $P((B \cup C) \mid A) = P(B \mid A) + P(C \mid A)$

Answer

1.

(a) The event $A$'s complementary event $\overline A$, in which you throw 4 times and get a 6 at least once, will be thrown 4 times and never get a 6's. Here, the probability of not having a 6 in each time is $${\frac{5}{6}}$$. Then

$$P_{r}(\overline A) = \left(\frac{5}{6}\right)^4$$

Therefore，

$$P_{r}(A) = 1 - P_{r}(\overline A) = 1 - \left(\frac{5}{6}\right)^4 = \frac{671}{1296}$$

(b) Consider the evemt $B$ in which two dice are thrown 24 times at the same time and both rolls 6 at least once. First, when you throw two dice at the same time, the probability that both will roll a 6 is $${\frac{1}{36}}$$．

Now the complement of the event $B$ is $\overline B$ and throw two dice 24 times at the same time, and both of them will not be 6 rolls. Thus,

$$P_{r}(\overline B) = \left(1 - \frac{1}{36}\right)^{24} = \left(\frac{35}{36}\right)^{24}$$

Therefore,

$$P_{r}(B) = 1 - P_{r}(\overline B) = 1 - \left(\frac{35}{36}\right)^{24} = 0.491$$

2 Let $A =$「true cancer patient」，$B =$「patients came out positive in precision inspection. Find the probability of being a cancer patient if the patient gives a positive result of the work-up．This can be expressed as follows using conditional probabilities.．

$$P_{r}(A \vert B)$$

Note that we know， $P_{r}(A) = 0.05$， $P_{r}(B \vert A) = 0.85$， $P_{r}(B \vert \overline A) = 0.05$． Then using，Bayes'theorem,

$$B = B \cap \Omega = B \cap (A \cup \overline A) = (B \cap A ) \cup (B \cap \overline A)$$

Thus

$$(1) \ P_{r}(B) = P_{r}(B \cap A) + P_{r}(B \cap \overline A) = P_{r}(A)P_{r}(B\vert A) + P_{r}(\overline A)P_{r}(B\vert\overline A)$$

$$= 0.05 \cdot 0.85 + 0.95 \cdot 0.05 = 0.9$$

$$(2) \ P_{r}(A \vert B) = \frac{P_{r}(A \cap B)}{P_{r}(B)} = \frac{P_{r}(A) P_{r}(B\vert A)}{P_{R}(B)} = \frac{P_{r}(A) P_{r}(B\vert A)}{P_{R}(B)}$$

$$= \frac{0.05 \cdot 0.85}{0.05 \cdot 0.85 + 0.95 \cdot 0.05} = \frac{0.425}{0.9} = 0.47$$

3.

(a)

$$x \in \overline{A \cup B} \Leftrightarrow x \not\in A \cup B$$

$$\Leftrightarrow x \not\in A \ {\rm and} \ x \not\in B$$

$$\Leftrightarrow x \in \overline A \ {\rm and} \ x \in \overline B$$

$$\Leftrightarrow x \in \overline A \cap x \in \overline B$$

Thus，

$$\overline{A \cup B} = \overline A \cap \overline B$$

Therefore，

$$P(\overline{A \cup B}) = P(\overline{A} \cap \overline{B})$$

(b)

$$x \in \overline{A \cap B} \Leftrightarrow x \not\in A \cap B$$

$$\Leftrightarrow x \not\in A \ {\rm or} \ x \not\in B$$

$$\Leftrightarrow x \in \overline A \ {\rm or} \ x \in \overline B$$

$$\Leftrightarrow x \in \overline A \cup x \in \overline B$$

Thus，

$$\overline{A \cap B} = \overline A \cup \overline B$$

Therefore，

$$P(\overline{A \cap B}) = P(\overline{A} \cup \overline{B})$$

(c) $B$ and $C$ are mutually exclusive. Then $B \mid A$ and $C \mid A$ are also mutually exclusive. Thus

$$P((B \cup C) \mid A) = P(B \mid A \cup C \mid A) = P(B \mid A) + P(C \mid A)$$