Probability

MATHEMATICS CLASS NOTES: PROBABILITY FOR SSS 2 Theoretical Frameworks To master the study of probability, we must first understand the mathematical models and philosophies that govern how we measure uncertainty. Probability is the branch of mathematics that quantifies the likelihood of an event occurring. Historically, this study evolved from games of chance in the 17th century, championed by mathematicians such as Blaise Pascal and Pierre de Fermat. Today, it forms the bedrock of statistics, insurance, quantum mechanics, and artificial intelligence. In modern mathematics, we analyze probability through three primary theoretical frameworks:

1. The Classical (A Priori) Approach This framework is applied when all possible outcomes of an experiment are known, finite, and equally likely to occur. It does not require physical experimentation; instead, it relies on deductive reasoning. Mathematically, if an experiment can result in NNN mutually exclusive, equally likely outcomes, and if nnn of these outcomes are favorable to an event AAA, then the classical probability of event AAA, denoted by P(A)P(A)P(A), is defined as: P(A)=n(A)n(S)=Number of favorable outcomesTotal number of possible outcomesP(A) = \frac{n(A)}{n(S)} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}P(A)=n(S)n(A)​=Total number of possible outcomesNumber of favorable outcomes​

Limitation: This approach fails when the total number of outcomes is infinite, or when the outcomes are not equally likely (for example, a biased coin or a weighted die).

1. The Empirical (Frequentist/Experimental) Approach When outcomes are not equally likely, or when the sample space is too complex to determine theoretically, we rely on observation and historical data. The empirical probability of an event is the limit of its relative frequency as the number of trials increases indefinitely. This is governed by the Law of Large Numbers. If an experiment is performed NNN times under identical conditions, and event AAA occurs fff times, the empirical probability of AAA is: P(A)=lim⁡N→∞fNP(A) = \lim_{N \to \infty} \frac{f}{N}P(A)=limN→∞​Nf​ In practical terms, for a sufficiently large number of trials NNN: P(A)≈Frequency of event ATotal number of trials NP(A) \approx \frac{\text{Frequency of event } A}{\text{Total number of trials } N}P(A)≈Total number of trials NFrequency of event A​

Example: If a factory tests 10,000 lightbulbs and finds 150 defective ones, the empirical probability of a bulb being defective is 15010000=0.015\frac{150}{10000} = 0.01510000150​=0.015 or 1.5%1.5%1.5%.

1. The Axiomatic Approach Developed by Andrey Kolmogorov in 1933, this is the modern, mathematically rigorous foundation of probability. It uses set theory to define probability without relying on physical experiments or the assumption of equally likely outcomes. The axiomatic framework states that probability is a real-valued function PPP that assigns a real number P(A)P(A)P(A) to every event AAA in a sample space SSS, satisfying three core axioms:

Axiom of Non-negativity: For any event AAA, P(A)≥0P(A) \geq 0P(A)≥0. Probability cannot be negative. Axiom of Certainty (Normalization): The probability of the entire sample space SSS is 111, i.e., P(S)=1P(S) = 1P(S)=1. Axiom of Additivity: For any sequence of mutually exclusive events A1,A2,A3,…A_1, A_2, A_3, \dotsA1​,A2​,A3​,… (where Ai∩Aj=∅A_i \cap A_j = \emptysetAi​∩Aj​=∅ for i≠ji \neq ji=j):

P(A1∪A2∪A3… )=P(A1)+P(A2)+P(A3)+…P(A_1 \cup A_2 \cup A_3 \dots) = P(A_1) + P(A_2) + P(A_3) + \dotsP(A1​∪A2​∪A3​…)=P(A1​)+P(A2​)+P(A3​)+… All mathematical properties and formulas of probability are logically derived from these three basic axioms.

In-Depth Core Concepts To build a solid foundation, we must define and analyze the core building blocks of probability. Experiment or Trial An experiment (or trial) is any systematic, repeatable process or action that generates a well-defined set of observable results. The key characteristic of a random experiment is that while the set of all possible results is known beforehand, the exact outcome of any single trial cannot be predicted with absolute certainty.

Example 1: Rolling a standard six-sided die is a random experiment. We know the possible results are numbers from 1 to 6, but we cannot predict which number will land face-up on the next roll. Example 2: Drawing a single card from a well-shuffled deck of 52 playing cards.

Outcome An outcome is a single, specific, and indivisible result obtained from running a single trial of an experiment.

Example: If you toss a coin once, "Heads" (HHH) is one outcome, and "Tails" (TTT) is another outcome. If you roll a die, getting a "444" is a single outcome.

Sample Space (SSS) The sample space (denoted by SSS or Ω\OmegaΩ) is the set containing all possible outcomes of a random experiment. It represents the universal set for that specific experiment. Defining the sample space accurately is the first and most critical step in solving any probability problem. Let us examine the sample spaces of common random experiments: A. Tossing Coins

One Coin: When a single coin is tossed, there are only two possible outcomes: Heads (HHH) or Tails (TTT). S={H,T},n(S)=2S = {H, T}, \quad n(S) = 2S={H,T},n(S)=2 Two Coins: When two coins are tossed simultaneously (or one coin is tossed twice), the sample space consists of ordered pairs, representing the outcome of the first coin and the second coin. S={(H,H),(H,T),(T,H),(T,T)},n(S)=4S = {(H,H), (H,T), (T,H), (T,T)}, \quad n(S) = 4S={(H,H),(H,T),(T,H),(T,T)},n(S)=4 Three Coins: S={(H,H,H),(H,H,T),(H,T,H),(H,T,T),(T,H,H),(T,H,T),(T,T,H),(T,T,T)},n(S)=8S = {(H,H,H), (H,H,T), (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T)}, \quad n(S) = 8S={(H,H,H),(H,H,T),(H,T,H),(H,T,T),(T,H,H),(T,H,T),(T,T,H),(T,T,T)},n(S)=8 Note: The general rule for tossing kkk coins is n(S)=2kn(S) = 2^kn(S)=2k.

B. Rolling Dice

One Die: A standard die has six faces numbered 1 to 6. S={1,2,3,4,5,6},n(S)=6S = {1, 2, 3, 4, 5, 6}, \quad n(S) = 6S={1,2,3,4,5,6},n(S)=6 Two Dice: When two distinct dice (e.g., a red die and a blue die) are rolled together, the sample space consists of ordered pairs (x,y)(x, y)(x,y), where xxx is the score on the first die and yyy is the score on the second die. This can be represented as a 6×66 \times 66×6 grid:

(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \ (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}$$

$n(S) = 6 \times 6 = 36$ Note: The general rule for rolling $k$ dice is $n(S) = 6^k$.

C. A Standard Deck of Playing Cards

A standard deck contains 52 cards, which are organized into four suits, two colors, and various ranks. Understanding this structure is essential for solving standard curriculum problems.

• Two Colors: Red (26 cards) and Black (26 cards).
• Four Suits (13 cards each):
  • Hearts ($\color{red}\heartsuit$): Red
  • Diamonds ($\color{red}\diamondsuit$): Red
  • Spades ($\spadesuit$): Black
  • Clubs ($\clubsuit$): Black
• Ranks within each suit: Ace (A), 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack (J), Queen (Q), King (K).
• Face Cards (Court Cards): Jack, Queen, and King. Since there are 3 face cards per suit, there are $3 \times 4 = 12$ face cards in a complete deck.
• Aces: There are 4 Aces in a deck (they are not typically classified as face cards).

Event ($E$)

An event is a subset of the sample space $S$. It consists of one or more outcomes that satisfy a specific condition. We denote events using capital letters (e.g., $A, B, C, E$).

• Simple Event: An event that contains exactly one outcome. For example, rolling a die and getting a "$6$". $E = \{6\}$, and $n(E) = 1$.
• Compound Event: An event that contains two or more outcomes. For example, rolling a die and getting an even number. Here, $E = \{2, 4, 6\}$, and $n(E) = 3$.

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Key Components in Detail

Now, let us analyze the relationships between different classes of events and the algebraic rules used to compute their probabilities.

1. Complementary Events

For any event $A$ defined on a sample space $S$, the complementary event of $A$ (denoted by $A'$, $A^c$, or $\bar{A}$) is the event that $A$ does not occur. It consists of all outcomes in the sample space $S$ that are not in $A$.

Using set theory notation: $A' = \{x \in S \mid x \notin A\}$

Mathematical Proof of Complementarity

Since $A$ and $A'$ are disjoint sets that together make up the entire sample space $S$:

1. $A \cap A' = \emptyset$ (They have no outcomes in common).
2. $A \cup A' = S$ (Their union is the entire sample space).

Applying the axiomatic probability framework: $P(A \cup A') = P(S)$ $P(A) + P(A') = 1 \quad \text{(Since } A \cap A' = \emptyset \text{ and } P(S) = 1\text{)}$

From this, we derive the fundamental formulas for complementary probability: $P(A') = 1 - P(A)$ $P(A) = 1 - P(A')$

Illustrative Example

Suppose a bag contains 5 red, 3 blue, and 2 green balls. A ball is drawn at random. Let $R$ be the event of drawing a red ball.

• Total outcomes: $n(S) = 5 + 3 + 2 = 10$.
• Favorable outcomes for $R$: $n(R) = 5$.
• $P(R) = \frac{5}{10} = 0.5$.

The complement of $R$ ($R'$) is the event of drawing a ball that is not red (i.e., blue or green).

• Method 1 (Direct): $n(R') = n(\text{Blue}) + n(\text{Green}) = 3 + 2 = 5$. Thus, $P(R') = \frac{5}{10} = 0.5$.
• Method 2 (Using Complement Rule): $P(R') = 1 - P(R) = 1 - 0.5 = 0.5$.

The complement rule is extremely useful when calculating the probability of "at least one" events, as it is often easier to compute the probability of "none" and subtract it from 1.

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2. Mutually Exclusive Events

Two events $A$ and $B$ are said to be mutually exclusive (or disjoint) if they cannot occur at the same time. If one event occurs, it prevents the other from occurring.

In terms of set theory, their intersection is an empty set: $A \cap B = \emptyset$ Therefore, the probability of their simultaneous occurrence is zero: $P(A \cap B) = 0$

The Addition Rule for Mutually Exclusive Events

When two events $A$ and $B$ are mutually exclusive, the probability that event $A$ or event $B$ occurs is the sum of their individual probabilities:

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

This is often referred to as the "OR" Rule.

Non-Mutually Exclusive Events (General Addition Rule)

If two events $A$ and $B$ can occur at the same time, they are non-mutually exclusive ($A \cap B \neq \emptyset$). If we simply add $P(A)$ and $P(B)$, we double-count the outcomes that belong to both events. To correct for this double-counting, we must subtract the probability of their intersection.

The General Addition Rule for any two events $A$ and $B$ is:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Comparison Table

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3. Independent Events

Two events $A$ and $B$ are independent if the occurrence (or non-occurrence) of one event has absolutely no effect on the probability of the occurrence of the other.

• Example: Tossing a coin and rolling a die. Getting a Head on the coin does not change the probability of rolling a $6$ on the die.

The Multiplication Rule for Independent Events

If $A$ and $B$ are independent events, the probability that both event $A$ and event $B$ occur is the product of their individual probabilities:

$P(A \cap B) = P(A) \times P(B)$

This is known as the "AND" Rule.

Generalization to Multiple Independent Events

If events $A_1, A_2, A_3, \dots, A_n$ are mutually independent, then: $P(A_1 \cap A_2 \cap A_3 \dots \cap A_n) = P(A_1) \times P(A_2) \times P(A_3) \times \dots \times P(A_n)$

Distinction: Mutually Exclusive vs. Independent Events

It is a common mistake to confuse mutually exclusive events with independent events.

• Mutually Exclusive means the events cannot happen together. If $A$ happens, $B$ definitely cannot happen. They are highly dependent on each other.
• Independent means the events have nothing to do with each other. The occurrence of $A$ provides zero information about whether $B$ will occur.

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4. Dependent Events and Conditional Probability

Two events are dependent if the occurrence of one event alters the probability of the other. This often occurs when we sample without replacement.

Conditional Probability

The conditional probability of an event $B$, given that event $A$ has already occurred, is denoted by $P(B \mid A)$ (read as "the probability of $B$ given $A$").

It is defined mathematically as: P(B \mid A) = \frac{P(A \cap B)}{P(A)} \quad \text{provided } P(A) > 0

From this definition, we can derive the General Multiplication Rule (which applies to all events, whether dependent or independent):

$P(A \cap B) = P(A) \times P(B \mid A)$

Note: If $A$ and $B$ are independent, $P(B \mid A) = P(B)$, and the formula simplifies back to $P(A \cap B) = P(A) \times P(B)$.

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Step-by-Step Solved Problems

To help you develop your problem-solving skills, let us work through several examples, moving from simple applications to more complex scenarios.

Problem 1: Rolling Two Dice (Sum and Properties)

Two fair six-sided dice are rolled simultaneously. Calculate the probability of obtaining:

1. A sum of exactly 8.
2. A sum of at least 10.
3. A double (the same number on both dice).

Solution:

First, define the total number of outcomes in the sample space $S$. Since each die has 6 outcomes, rolling two dice gives: $n(S) = 6 \times 6 = 36$

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Part 1: Probability of getting a sum of exactly 8.

Let $A$ be the event that the sum of the two faces is 8. Identify the ordered pairs $(x, y)$ in the $6 \times 6$ grid that sum to 8: $A = \{(2,6), (3,5), (4,4), (5,3), (6,2)\}$ Count the number of favorable outcomes: $n(A) = 5$ Compute the probability: $P(A) = \frac{n(A)}{n(S)} = \frac{5}{36}$

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Part 2: Probability of getting a sum of at least 10.

"At least 10" means the sum can be 10, 11, or 12. Let $B$ be this event. List the outcomes for each acceptable sum:

• Sum of 10: $\{(4,6), (5,5), (6,4)\}$
• Sum of 11: $\{(5,6), (6,5)\}$
• Sum of 12: $\{(6,6)\}$

Combine these to write out event $B$: $B = \{(4,6), (5,5), (6,4), (5,6), (6,5), (6,6)\}$ Count the outcomes: $n(B) = 6$ Compute the probability: $P(B) = \frac{n(B)}{n(S)} = \frac{6}{36} = \frac{1}{6}$

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Part 3: Probability of getting a double.

Let $C$ be the event of obtaining a double (the same number on both dice). Identify the outcomes: $C = \{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}$ Count the outcomes: $n(C) = 6$ Compute the probability: $P(C) = \frac{n(C)}{n(S)} = \frac{6}{36} = \frac{1}{6}$

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Problem 2: Card Probability (Addition Rule Application)

A single card is drawn at random from a standard, well-shuffled deck of 52 playing cards. Calculate the probability that the drawn card is:

1. A King or a Queen.
2. A Red card or a Jack.

Solution:

The total number of outcomes in the sample space is: $n(S) = 52$

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Part 1: Probability of drawing a King or a Queen.

Let $K$ be the event of drawing a King, and $Q$ be the event of drawing a Queen.

• Number of Kings in a deck: $n(K) = 4 \implies P(K) = \frac{4}{52}$
• Number of Queens in a deck: $n(Q) = 4 \implies P(Q) = \frac{4}{52}$

Are these events mutually exclusive? Yes, because a single card cannot be both a King and a Queen at the same time ($K \cap Q = \emptyset$).

Apply the addition rule for mutually exclusive events: $P(K \cup Q) = P(K) + P(Q)$ $P(K \cup Q) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52} = \frac{2}{13}$

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Part 2: Probability of drawing a Red card or a Jack.

Let $R$ be the event of drawing a red card, and $J$ be the event of drawing a Jack.

• Number of Red cards: $n(R) = 26 \implies P(R) = \frac{26}{52}$
• Number of Jacks: $n(J) = 4 \implies P(J) = \frac{4}{52}$

Are these events mutually exclusive? No, because there are cards that are both red and Jacks (the Jack of Hearts and the Jack of Diamonds). $n(R \cap J) = 2 \implies P(R \cap J) = \frac{2}{52}$

Apply the general addition rule for non-mutually exclusive events: $P(R \cup J) = P(R) + P(J) - P(R \cap J)$ $P(R \cup J) = \frac{26}{52} + \frac{4}{52} - \frac{2}{52}$ $P(R \cup J) = \frac{26 + 4 - 2}{52} = \frac{28}{52} = \frac{7}{13}$

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Problem 3: Independent Events (With Replacement)

A bag contains 6 black marbles and 4 white marbles. A marble is drawn at random, its color is recorded, and it is placed back into the bag (with replacement). A second marble is then drawn. Calculate the probability that:

1. Both marbles are black.
2. The first marble is black and the second is white.
3. The marbles are of different colors.

Solution:

• Total number of marbles: $6 + 4 = 10$.
• Since the marble is replaced before the second draw, the total number of marbles remains 10 for both trials. The outcomes of the first and second draws are independent.
• Probability of drawing a Black marble on any draw: $P(B) = \frac{6}{10} = \frac{3}{5}$
• Probability of drawing a White marble on any draw: $P(W) = \frac{4}{10} = \frac{2}{5}$

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Part 1: Probability that both marbles are black.

This is the event of drawing a Black marble first and a Black marble second. Since the events are independent, apply the multiplication rule: $P(B_1 \cap B_2) = P(B) \times P(B)$ $P(B_1 \cap B_2) = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25}$

In decimal form: $0.36$ or $36\%$.

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Part 2: Probability that the first marble is black and the second is white.

Apply the multiplication rule: $P(B_1 \cap W_2) = P(B) \times P(W)$ $P(B_1 \cap W_2) = \frac{3}{5} \times \frac{2}{5} = \frac{6}{25}$

In decimal form: $0.24$ or $24\%$.

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Part 3: Probability that the marbles are of different colors.

There are two mutually exclusive ways to draw two different colored marbles:

• Case 1: First is Black, second is White ($B_1 \cap W_2$)
• Case 2: First is White, second is Black ($W_1 \cap B_2$)

First, calculate the probability of Case 2: $P(W_1 \cap B_2) = P(W) \times P(B) = \frac{2}{5} \times \frac{3}{5} = \frac{6}{25}$

Since Case 1 and Case 2 are mutually exclusive (you cannot have both sequences happen in the same two-draw trial), we add their probabilities: $P(\text{Different Colors}) = P(B_1 \cap W_2) + P(W_1 \cap B_2)$ $P(\text{Different Colors}) = \frac{6}{25} + \frac{6}{25} = \frac{12}{25}$

In decimal form: $0.48$ or $48\%$.

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Problem 4: Dependent Events (Without Replacement)

Let us use the same bag of marbles (6 black, 4 white). This time, a marble is drawn, but it is not replaced before drawing the second marble. Calculate the probability that:

1. Both marbles are black.
2. The first marble is black and the second is white.
3. At least one white marble is drawn.

Solution:

Because we are drawing without replacement, the total number of marbles in the bag changes from 10 to 9 on the second draw, and the number of marbles of the color we drew first decreases by 1. These are dependent events.

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Part 1: Probability that both marbles are black.

• First draw: 6 black marbles out of 10. $P(B_1) = \frac{6}{10} = \frac{3}{5}$
• Second draw: If the first marble drawn was black, there are now 5 black marbles left, and 9 total marbles left in the bag. $P(B_2 \mid B_1) = \frac{5}{9}$

Apply the multiplication rule for dependent events: $P(B_1 \cap B_2) = P(B_1) \times P(B_2 \mid B_1)$ $P(B_1 \cap B_2) = \frac{6}{10} \times \frac{5}{9} = \frac{30}{90} = \frac{1}{3}$

In decimal form: $\approx 0.333$ or $33.3\%$.

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Part 2: Probability that the first is black and the second is white.

• First draw: 6 black marbles out of 10. $P(B_1) = \frac{6}{10}$
• Second draw: If a black marble was drawn first, all 4 white marbles are still in the bag, but the total number of marbles is now 9. $P(W_2 \mid B_1) = \frac{4}{9}$

Apply the formula: $P(B_1 \cap W_2) = P(B_1) \times P(W_2 \mid B_1)$ $P(B_1 \cap W_2) = \frac{6}{10} \times \frac{4}{9} = \frac{24}{90} = \frac{4}{15}$

In decimal form: $\approx 0.267$ or $26.7\%$.

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Part 3: Probability of drawing at least one white marble.

The event "at least one white marble" can occur in three ways: (Black, White), (White, Black), or (White, White).

While we can calculate each of these and add them together, it is much faster to use the complement rule. The complement of "at least one white" is "no white marbles at all" (which means both marbles must be black).

Using the complement rule: $P(\text{At least one White}) = 1 - P(\text{No White}) = 1 - P(B_1 \cap B_2)$

We calculated $P(B_1 \cap B_2)$ in Part 1 as $\frac{1}{3}$. $P(\text{At least one White}) = 1 - \frac{1}{3} = \frac{2}{3}$

In decimal form: $\approx 0.667$ or $66.7\%$.

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Problem 5: Complex Algebraic System

In a class of SSS2 students, the probability that a randomly chosen student passes a Mathematics test is $0.8$, and the probability that they pass an English test is $0.7$. If the probability that they pass both tests is $0.6$, calculate the probability that a student chosen at random:

1. Passes Mathematics or English.
2. Passes Mathematics but fails English.
3. Fails both tests.

Solution:

Let:

• $M$ be the event of passing Mathematics. Given: $P(M) = 0.8$.
• $E$ be the event of passing English. Given: $P(E) = 0.7$.
• $M \cap E$ be the event of passing both. Given: $P(M \cap E) = 0.6$.

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Part 1: Probability that a student passes Mathematics or English.

Since $P(M \cap E) = 0.6 \neq 0$, these events are non-mutually exclusive. Apply the general addition rule: $P(M \cup E) = P(M) + P(E) - P(M \cap E)$ $P(M \cup E) = 0.8 + 0.7 - 0.6$ $P(M \cup E) = 1.5 - 0.6 = 0.9$

The probability that a student passes at least one of the subjects is $0.9$.

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Part 2: Probability that a student passes Mathematics but fails English.

This event is represented as $M \cap E'$ (passing Mathematics and not passing English). We can find this using a Venn diagram or the following formula: $P(M \cap E') = P(M) - P(M \cap E)$

Intuition: The probability of passing Math consists of two parts: passing Math and English, or passing Math and failing English. Subtracting the overlapping part gives us the probability of passing only Math.

$P(M \cap E') = 0.8 - 0.6 = 0.2$

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Part 3: Probability that a student fails both tests.

Failing both tests means a student does not pass Math and does not pass English. This is represented as $M' \cap E'$. Using De Morgan's Laws from set theory, we know that: $M' \cap E' = (M \cup E)'$

This tells us that the event of failing both is the complement of passing at least one of the subjects. Applying the complement rule: $P(M' \cap E') = 1 - P(M \cup E)$

We calculated $P(M \cup E) = 0.9$ in Part 1. $P(M' \cap E') = 1 - 0.9 = 0.1$

The probability of failing both subjects is $0.1$ or $10\%$.

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Comprehensive Summary

To round out your understanding of SSS2 probability, let us review the key definitions and formulas in this summary guide.

Core Terminology Cheat Sheet

• Probability ($P$): A numerical measure of the likelihood of an event occurring, bounded such that $0 \le P(A) \le 1$.
• Sample Space ($S$): The complete set of all possible outcomes for a given experiment.
• Complement ($A'$): The event containing all outcomes in the sample space that are not in event $A$.
• Mutually Exclusive: Events that cannot occur at the same time ($P(A \cap B) = 0$).
• Independent: Events where the occurrence of one does not affect the probability of the other ($P(A \cap B) = P(A) \times P(B)$).

Summary of Key Probability Formulas