Basic Probability Rules and Counting Methods

Elementary Event Rule

• Let $S$ represent the sample space such that $S = \{e_1, e_2, \dots, e_n\}$.

• Let $E$ represent an event within the sample space such that $E = \{e_1, e_2, \dots, e_m\}$.

• The sum of probabilities of all elementary events in the sample space must equal 1: $P(S) = P(e_1) + P(e_2) + \dots + P(e_n) = 1$

• The probability of a specific event $E$ is the sum of the probabilities of the outcomes that compose it:
  $P(E) = P(e_1) + P(e_2) + \dots + P(e_m) = \frac{n(E)}{n(S)}$

Addition Rule for Two Events

• For any two events $E$ and $F$ in the sample space $S$:

  • $E \cap F$ denotes the intersection of events $E$ and $F$.

  • $E \cup F$ denotes the union of events $E$ and $F$.

• There are two cases for computing the probability of the union, $P(E \cup F)$, depending on whether the events overlap.

Case 1: Mutually Exclusive Events

• Events $E$ and $F$ are mutually exclusive if they cannot occur at the same time ($E \cap F = \emptyset$).

• If $n(S) = n$, $n(E) = n_1$, and $n(F) = n_2$, then: $P(E \cup F) = \frac{n_1 + n_2}{n}$

• Therefore: $P(E \cup F) = P(E) + P(F)$

Case 2: Non-Mutually Exclusive Events

• If events $E$ and $F$ are not mutually exclusive, they share common elements.

• The addition rule is: $P(E \cup F) = P(E) + P(F) - P(E \cap F)$

• Numerical Example:

  • Sample space $S = \{a, b, c, d, e, f, g\}$, so $n(S) = 7$.

  • Event $E = \{c, d, e\}$, so $n(E) = 3$.

  • Event $F = \{a, b, c\}$, so $n(F) = 3$.

  • Intersection $E \cap F = \{c\}$, so $n(E \cap F) = 1$.

  • Union $E \cup F = \{a, b, c, d, e\}$, so $n(E \cup F) = 5$.

  • Using the rule: $P(E \cup F) = P(E) + P(F) - P(E \cap F) = \frac{3}{7} + \frac{3}{7} - \frac{1}{7} = \frac{5}{7}$.

Complement Rule

• Let $G$ be an event in the sample space $S$.

• The complement of event $E$ is denoted as $\bar{E}$.

• The intersection of an event and its complement is an empty set: $E \cap \bar{E} = \emptyset$

Conditional Event Rule

• For any two sets (events) $E$ and $F$, the notation $E/F$ (read as $E$ given $F$) represents the probability that event $E$ occurs given that event $F$ has already occurred.

• If $P(F) \neq 0$, then: $P(E|F) = \frac{P(E \cap F)}{P(F)}$

• Conversely, if $P(E) \neq 0$, then: $P(F|E) = \frac{P(E \cap F)}{P(E)}$

• Generally, for any events $x$ and $y$: $P(x|y) = \frac{P(x \cap y)}{P(y)}$

Independent Events

• Two events $E$ and $F$ are considered independent if the occurrence of one does not affect the occurrence of the other.

• The probability of their intersection is the product of their individual probabilities: $P(E \cap F) = P(E) \times P(F)$

Partitioned Event Rule

• Consider a population divided into $k$ various partitions (sub-populations):

  • Subpop 1 contains $N_1$ items.

  • Subpop 2 contains $N_2$ items.

  • Subpop $k$ contains $N_k$ items.

  • The total population is $N = N_1 + N_2 + \dots + N_k$.

• Suppose a sample of size $n$ is taken, consisting of:

  • $n_1$ items from Subpop 1.

  • $n_2$ items from Subpop 2.

  • $n_k$ items from Subpop $k$.

  • The total sample size is $n = n_1 + n_2 + \dots + n_k$.

• Let $E_i$ be the event of selecting $n_i$ items from sub-population $i$ for all $i = 1, 2, \dots, k$.

• The probability $P(E)$ is calculated using permutations or combinations ($C$): $P(E) = \frac{n(E)}{n(S)} = \frac{\binom{N_1}{n_1} \times \binom{N_2}{n_2} \times \dots \times \binom{N_k}{n_k}}{\binom{N}{n}}$

Counting Methods

Multiplication Principle

• Let $E_1$ and $E_2$ be any two events of interest.

• Let $n_1 = n(E_1)$ be the number of occurrences of $E_1$.

• Let $n_2 = n(E_2)$ be the number of occurrences of $E_2$.

• The total number of ways both $E_1$ and $E_2$ can occur is: $n_1 \times n_2 = n(E_1) \times n(E_2)$

The Factorial Formula

• Consider an experiment involving the assignment of $n$ students to $n$ projects.

• Let $E_i$ be the event of assigning project $i$ to student $i$ for $i = 1, 2, \dots, n$.

• The number of ways of assigning project $i$ ($n(E_i)$) decreases as slots are filled:

  • $n(E_1) = n$

  • $n(E_2) = n - 1$

  • $n(E_3) = n - 2$

  • $n(E_n) = 1$

• For example, assigning 10 students to projects results in $10 \times 9 \times 8 \times 7 \times \dots \times 1$

• The total number of ways of assigning $n$ students to $n$ projects is: $n(E_1) \times n(E_2) \times n(E_3) \times \dots \times n(E_n) = n \times (n-1) \times (n-2) \times (n-3) \times \dots \times 1 = n!$

• $n!$ is read as "$n$ factorial."

Permutations and Combinations

Permutation

• Consider an experiment of assigning $n$ students to $r$ projects where r < n.

• The number of ways of assigning projects is given by the permutation: $P(n, r) = n(E_1) \times n(E_2) \times \dots \times n(E_r)$

• This simplifies to the formula: $P(n, r) = \frac{n!}{(n - r)!}$

• A permutation is defined as the number of ways items can be arranged in a given order.

Combination

• If the order of arranging items is not important, the event is called a Combination ($C(n, r)$ or $^nC_r$).

• The formula is: $C(n, r) = \frac{n!}{r! (n - r)!} = \frac{P(n, r)}{r!}$

Permutation of Non-Distinct Items

• Consider an experiment where $n$ students are assigned to $n$ projects that are partitioned into $k$ groups.

• The projects are distributed as follows:

  • $n_1$ projects are of type 1.

  • $n_2$ projects are of type 2.

  • $n_k$ projects are of type $k$.

  • The sum of all types equals the total: $n_1 + n_2 + \dots + n_k = n$.

• If $E_i$ is the event of assigning $n_i$ projects of type $i$ to students, the number of ways to perform this assignment is: $P(n_1, n_2, n_3, \dots, n_k) = \frac{n!}{n_1! \times n_2! \times n_3! \times \dots \times n_k!}$