Fundamentals of Probability

Probability is defined as the likelihood that a specific random event will occur in the future. To understand this field, we must first establish a common vocabulary of its core components.

I. Basic Terminology

Experiment: An act that can be repeated under specific conditions where the outcome is not certain. For example, when throwing a die, the possible outcomes are $1, 2, 3, 4, 5, 6$ , but the exact result of any single throw cannot be predicted.
Event: One or more specified outcomes from an experiment. For instance, in a coin toss, "Heads" (H) occurring is an event.
Compound Event: Occurs when two or more simple events happen simultaneously or in connection with each other. If A and B are two simple events, their joint occurrence is denoted as $(A B)$ .
Sample Space (S): The set containing the totality of all possible outcomes for an experiment. For a coin toss, $S = H, T$ .

II. Classifying Events

Mutually Exclusive: Two or more events that cannot occur at the same time. The occurrence of one automatically excludes the others.
Independent: Events are independent if the occurrence of one does not influence or affect the occurrence of the other. Mathematically, two events are independent if: $$P(AB) = P(A) \times P(B)$$
Equally Likely: Events that have an equal chance of occurring.
Exhaustive: A set of events is exhaustive if they include all possible outcomes of an experiment.
Conditional: When the occurrence of one event depends on whether another event has already occurred. This is denoted as $P (A / B)$ , meaning the probability of event A occurring given that event B has occurred.

III. Definitions of Probability

There are two primary ways to define and calculate probability:

Mathematical (Classical) Definition: Used when a trial results in $n$ exhaustive, mutually exclusive, and equally likely cases. If $m$ of those cases are favorable to event A, then: $$P(A) = \frac{\text{Favorable outcomes (m)}}{\text{Total outcomes (n)}}$$
- Limitations: This cannot be used if outcomes are not equally likely or if the number of possible outcomes is infinite.
Statistical (Empirical) Definition: Based on the "Relative Frequency" of an event. If event A occurs $r$ times in $n$ independent trials, the probability is the limit of the ratio $r / n$ as $n$ approaches infinity: $$P(A) = \lim_{n \to \infty} \frac{r}{n}$$
- Example: In 1,000 coin tosses, if "Heads" appears 500 times, the empirical probability is $500 / 1000 = 0.50$ .

IV. The Laws of Probability

1. The Additive Law (Theorem of Total Probability)

Statement: For two events that are not mutually exclusive, the probability that at least one of them occurs is the sum of their individual probabilities minus the probability of them happening together. Formula: $P (A_{1} + A_{2}) = P (A_{1}) + P (A_{2}) - P (A_{1} A_{2})$

Proof (Step-by-Step):

Imagine a Venn diagram where two circles (Events $A_{1}$ and $A_{2}$ ) overlap.
The area of $A_{1}$ is composed of the part that doesn't overlap with $A_{2}$ (denoted $A_{1} \bar{A_{2}}$ ) and the overlapping part ( $A_{1} A_{2}$ ). Thus: $P (A_{1}) = P (A_{1} \bar{A_{2}}) + P (A_{1} A_{2})$ (Equation i).
Similarly, for $A_{2}$ : $P (A_{2}) = P (\bar{A_{1}} A_{2}) + P (A_{1} A_{2})$ (Equation ii).
The total area covered by both circles ( $A_{1} + A_{2}$ ) is the sum of the three distinct sections: $P (A_{1} + A_{2}) = P (A_{1} \bar{A_{2}}) + P (A_{1} A_{2}) + P (\bar{A_{1}} A_{2})$ (Equation iii).
If we add Equation (i) and Equation (ii) together, we get the sum of all four sections. By subtracting the overlapping part ( $P (A_{1} A_{2})$ ) once, we arrive at the total area defined in Equation (iii).

Special Cases:

If events are Independent: $P (A_{1} + A_{2}) = P (A_{1}) + P (A_{2}) - [P (A_{1}) \times P (A_{2})]$ .
If events are Mutually Exclusive: Since they can't overlap, $P (A_{1} A_{2}) = 0$ , so $P (A_{1} + A_{2}) = P (A_{1}) + P (A_{2})$ .

2. The Multiplicative Law (Theorem of Compound Probability)

Statement: For two dependent events, the probability of both occurring is the probability of the first event multiplied by the conditional probability of the second. Formula: $P (A B) = P (A) \times P (B / A)$

Proof (Step-by-Step):

Let $n$ be the total outcomes, $n_{1}$ be outcomes favorable to A, $n_{2}$ favorable to B, and $n_{12}$ favorable to both.
By definition: $P (A) = n_{1} / n$ and $P (B / A) = n_{12} / n_{1}$ (Probability of both occurring, out of only those where A occurred).
Multiply them: $(n_{1} / n) \times (n_{12} / n_{1}) = n_{12} / n$ . This equals the definition of $P (A B)$ .

V. Practical Problem Solving

Problem 1: Numbered Tickets There are 20 tickets (1–20). Find the probability that a random ticket is a multiple of 2 or 5.

Step 1: List multiples of 2: $2, 4, 6, 8, 10, 12, 14, 16, 18, 20$ . There are 10. $P (A) = 10 / 20$ .
Step 2: List multiples of 5: $5, 10, 15, 20$ . There are 4. $P (B) = 4 / 20$ .
Step 3: Identify overlapping numbers (multiples of both 2 and 5): $10, 20$ . There are 2. $P (A B) = 2 / 20$ .
Step 4: Use Additive Law: $10 / 20 + 4 / 20 - 2 / 20 = 12 / 20$ . Simplified, the answer is 3/5.

Problem 2: Team Selection (Combinations) A team of 4 must be formed from 4 accountants, 2 economists, 3 statisticians, and 1 physician (10 total employees).

Total Outcomes: We choose 4 out of 10. Using combinations ( $^{10} C_{4}$ ): $$\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \text{ ways}$$
Task: Probability the team includes at least 1 statistician.
- Step 1: Calculate the probability of the opposite (no statisticians).
- Step 2: If we exclude the 3 statisticians, we choose 4 people from the remaining 7 employees ( $^{7} C_{4}$ ): $$\frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1} = 35 \text{ ways}$$
- Step 3: Probability of NO statisticians = $35 / 210 = 1 / 6$ .
- Step 4: Probability of AT LEAST one = $1 - 1 / 6 = 5 / 6$ .

Problem 3: Drawing Balls from an Urn An urn has 8 white, 6 red, and 3 blue balls (17 total). Two are drawn. Find the probability they are the same color.

Total Outcomes: Choosing 2 from 17 ( $^{17} C_{2}$ ): $$\frac{17 \times 16}{2 \times 1} = 136 \text{ ways}$$
Favorable Outcomes (Same Color):
- Two White ( $^{8} C_{2}$ ): $\frac{8 \times 7}{2 \times 1} = 28$
- Two Red ( $^{6} C_{2}$ ): $\frac{6 \times 5}{2 \times 1} = 15$
- Two Blue ( $^{3} C_{2}$ ): $\frac{3 \times 2}{2 \times 1} = 3$
Total Favorable: $28 + 15 + 3 = 46$ .
Probability: $46 / 136 = 23 / 68$ .