I am writing some notes on basic probability in my own version, but regarding to many different versions of probability concepts I have found in several books and articles..I would like to ask whether or not there is a confusion/mistake in my version. Thanks.
*This writing is done without looking at some books for references or even copying, but by past experience reading math books and studying math. Except for the opening "Probability is the measure of the chance that an event will occur." is from Wikipedia.
Here are my writings:
Probability is the measure of the chance that an event will occur. To illustrate probability, we start with a classic example. Suppose there is a box containing 3 red balls, and 2 blue balls. Let the balls be denoted as $r_{1}, r_{2}, r_{3}, b_{1},$ and $b_{2}$, and let $B = \{r_{1}, r_{2}, r_{3}, b_{1}, b_{2}\}$. Then suppose you pick a ball randomly from the box without looking at the box. Because of random picking, then each of the 5 balls has the same chance to be picked. The probability that the ball $x \in B$ will be the picked one is the numerical value for this chance, in this case it will be $1/5$. Why $1/5$? here are the rules of probability (using the random ball pick for illustration of some of the rules):
- The random ball pick is an experiment.
- The set of all possible outcomes of the experiment is called the sample space, let this be $\Omega$. In this example we have the total number of possible outcomes is 5. For this example, let $R_{i}$ be the outcome that ball $r_{i}$ is picked, and let $B_{i}$ be the outcome that ball $b_{i}$ is picked. Now notice that $\Omega = \{R_{1}, R_{2}, R_{3}, B_{1}, B_{2} \}$.
- Probability ranges from $0-1$, a probability of $0$ means that it is impossible, and $1$ means that it is certain.
- $P(x)$ means the probability of an event $x$ occurs. Notice that $0 \le P(x) \le 1, \forall x \in \Omega$ must holds. If each outcome in $\Omega$ has the same chance, then the probability for each outcome is defined as $$ \frac{1}{|\Omega|} $$ (this explain the number $1/5$), and the probability of an event is defined as $$ \frac{|\Omega'|}{|\Omega|} $$ where $\Omega' \subseteq \Omega$ is the set of all outcomes corresponding to the event. Here are some examples regarding to this point, 1) The probability of getting a red ball will be $3/5$, because there are 3 outcomes corresponding to the event, which are in $\{R_{1}, R_{2}, R_{3}\}$. It makes sense that this is larger than the probability of getting a red ball $r_{1}$ which will be just $1/5$, because the event getting a red ball offers more possibilities (more likely tho happen) than the stricter event getting a red ball $r_{1}$. 2) The probability of getting a red ball and a blue ball will be $0/5 = 0$, because there is no outcome corresponding to the event. This is logical, it is not possible to picked one ball and expect it to be both red and blue. 3) The probability of getting a ball (does not matter the color) will be $5/5=1$, which makes sense because it means that this event will certainly happen.
- In general, each subset $\omega \subseteq \Omega$ is an event. If a subset has more than one elements, then it is interpreted as the event of an outcome in the subset occurs. For example, $\{R_{1}, R_{2}\}$ is the event that a red ball is picked (either $r_{1}$ or $r_{2}$ is picked), we can see this set as $R_{1} \cup R_{2}$. Another example, $\{R_{1}, R_{2}, B_{1}\}$ is the event that either $r_{1}$ or $r_{2}$ or $b_{1}$ is picked, we can see this set as $R_{1} \cup R_{2} \cup B_{1}$.
- In general, for any two different outcomes $x, y \in \Omega$, we must have $$ P(x \cup y) = P(x) + P(y)$$.
- In general, for any two different event $\omega_{1}, \omega_{2} \subseteq \Omega$, we must have $$ P(\omega_{1} \cup \omega_{2}) = P(\omega_{1}) + P(\omega_{2}) - P(\omega_{1} \cap \omega_{2})$$ *this rule does not look clear for this example, but it holds in general.
Next, let us improve the example/experiment by considering the case where you pick 2 balls randomly from the box, at one take. The possibilities can be listed as these sets: $$ \Omega = \{ p_{r_{1}, r_{2}}, p_{r_{1}, r_{3}}, p_{r_{1}, b_{1}}, p_{r_{1}, b_{2}}, p_{r_{2}, r_{3}}, p_{r_{2}, b_{1}}, p_{r_{2}, b_{2}}, p_{r_{3}, b_{1}}, p_{r_{3}, b_{2}}, p_{b_{1}, b_{2}} \} $$, there are exactly $\binom{5}{2} = 10$ possible outcomes. Since each event has the same chance of being picked, then each will have probability of $1/10$. Some facts for this example: $P(p_{r_{1}, r_{2}}) = 1/10$, $P(p_{r_{1}, r_{2}} \cup p_{r_{2}, r_{3}}) = 1/10 + 1/10 = 1/5$, the probability of getting at least one blue ball is the same as $P( p_{r_{1}, b_{1}} \cup p_{r_{1}, b_{2}} \cup p_{r_{2}, b_{1}} \cup p_{r_{2}, b_{2}} \cup p_{r_{3}, b_{1}} \cup p_{r_{3}, b_{2}} \cup p_{b_{1}, b_{2}} ) = 7 \cdot 1/10 = 7/10$.
We can also see a probability value as relative frequency to the sample space. In the last example, there are 7 possible events that correspond to teh event $\omega:= \text{"at least one blue ball is picked"}$ out of all 10 possible events, so the probability $P(\omega)=7/10$. This can only hold when all the events in $\Omega$ have equal chance.