🎲 Chapter 13: Probability

The conditional probability of event E, given that event F has already occurred, is denoted P(E|F) and defined as:

P(E|F) = P(E ∩ F) / P(F), provided P(F) ≠ 0

Properties of conditional probability:

• 0 ≤ P(E|F) ≤ 1
• P(S|F) = P(F|F) = 1
• P((A∪B)|F) = P(A|F) + P(B|F) − P((A∩B)|F)
• P(E′|F) = 1 − P(E|F)

Two events E and F are said to be independent if the occurrence of one does not affect the probability of occurrence of the other:

P(E ∩ F) = P(E) · P(F)

Equivalently, P(E|F) = P(E) and P(F|E) = P(F), provided the conditional probabilities are defined.

Note: Independent events and mutually exclusive events are different concepts — two events with nonzero probabilities cannot be both mutually exclusive and independent.

If E₁, E₂, ..., Eₙ are mutually exclusive and exhaustive events with nonzero probabilities forming a partition of sample space S, and A is any event with P(A) > 0, then:

P(Eᵢ|A) = [P(Eᵢ)·P(A|Eᵢ)] / [Σ P(Eⱼ)·P(A|Eⱼ)]

Here, P(Eᵢ) are called priori probabilities (known before the experiment), and P(Eᵢ|A) are called posteriori probabilities (computed after event A has occurred). Bayes' theorem is widely used to "reverse" conditional probabilities.

Random variable: A real-valued function whose domain is the sample space of a random experiment. Usually denoted X.

Probability distribution: A description that gives the values a random variable can take, along with their corresponding probabilities. If X takes values x₁, x₂, ..., xₙ with probabilities p₁, p₂, ..., pₙ, then:

• pᵢ > 0 for all i
• Σ pᵢ = 1 (sum of all probabilities is 1)

Bernoulli trials: A sequence of trials is called Bernoulli trials if:

• There are a finite number of trials.
• The trials are independent of each other.
• Each trial has exactly two outcomes: success or failure.
• The probability of success remains the same in every trial.

Binomial distribution: For n Bernoulli trials with probability of success p and failure q = 1 − p, the probability of exactly x successes is:

P(X = x) = ⁿCₓ · pˣ · qⁿ⁻ˣ, x = 0, 1, 2, ..., n

Mean = np, Variance = npq

✏️ Example 1: A die is thrown. Find P(number is even | number is greater than 3).

Solution: Let E = even number = {2,4,6}, F = number > 3 = {4,5,6}

E ∩ F = {4,6}, so P(E∩F) = 2/6, P(F) = 3/6

P(E|F) = (2/6)/(3/6) = 2/3

✏️ Example 2: Bag A has 3 red, 2 black balls; Bag B has 2 red, 4 black balls. A bag is chosen at random and a ball drawn — it turns out red. Find the probability it came from Bag A.

Solution: P(A) = P(B) = 1/2

P(Red|A) = 3/5, P(Red|B) = 2/6 = 1/3

By Bayes' theorem: P(A|Red) = [P(A)P(Red|A)] / [P(A)P(Red|A) + P(B)P(Red|B)]

= (1/2 × 3/5) / [(1/2×3/5) + (1/2×1/3)] = (3/10) / (3/10 + 1/6)

= (3/10) / (14/30) = (9/30)/(14/30) = 9/14

✏️ Example 3: A coin is tossed 4 times. Find the probability of getting exactly 3 heads.

Solution: n = 4, p = 1/2, q = 1/2, x = 3

P(X=3) = ⁴C₃ (1/2)³(1/2)¹ = 4 × (1/16) = 4/16

P(X=3) = 1/4