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Probability - Axiomatic Approach to Probability

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Random Experiment and Sample Space: A random experiment is a process where the outcome cannot be predicted with certainty, but the set of all possible outcomes is known. This set is called the Sample Space, denoted by SS. Visually, imagine SS as a rectangular boundary containing all possible individual outcomes (sample points).

Events: An event is a subset of the sample space SS. An event AA is said to have occurred if the outcome of the experiment is an element of set AA. In a Venn diagram, an event is depicted as a circle or a closed region inside the sample space rectangle.

Axiomatic Definition of Probability: Let SS be the sample space. Probability PP is a function that assigns a real number to each event such that: (i) The probability of any event is non-negative (P(A)0P(A) \ge 0); (ii) The probability of the entire sample space is 1 (P(S)=1P(S) = 1); (iii) For mutually exclusive events, the probability of their union is the sum of their individual probabilities.

Mutually Exclusive Events: Two events AA and BB are mutually exclusive if the occurrence of one excludes the occurrence of the other, meaning AB=ϕA \cap B = \phi. Visually, these are represented as two separate circles within the sample space that do not overlap or share any area.

Exhaustive Events: A set of events E1,E2,...,EnE_1, E_2, ..., E_n are exhaustive if their union covers the entire sample space, i.e., E1E2...En=SE_1 \cup E_2 \cup ... \cup E_n = S. Visually, this means that the combined area of these events fills the entire sample space rectangle.

Complementary Events: For any event AA, the event 'not AA', denoted by AA' or AcA^c, contains all outcomes in SS that are not in AA. Visually, if AA is a circular region, AA' is the entire area of the sample space rectangle that lies outside that circle.

Equally Likely Outcomes: If a sample space has nn outcomes and each has the same chance of occurring, the probability of each elementary event is 1n\frac{1}{n}. If an event AA consists of mm such outcomes, then P(A)=mnP(A) = \frac{m}{n}.

Algebra of Events: This involves operations like Union (ABA \cup B, representing 'A or B' or 'at least one'), Intersection (ABA \cap B, representing 'A and B' or 'simultaneous occurrence'), and Difference (ABA - B, representing 'A but not B'). Visually, ABA \cup B is the total area covered by both circles, while ABA \cap B is the overlapping shaded region.

📐Formulae

0P(A)10 \le P(A) \le 1

P(S)=1P(S) = 1

P(ϕ)=0P(\phi) = 0

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

P(A)=1P(A)P(A') = 1 - P(A)

P(AB)=P(A)+P(B) (for mutually exclusive events)P(A \cup B) = P(A) + P(B) \text{ (for mutually exclusive events)}

P(AB)=P(A)P(AB)P(A \cap B') = P(A) - P(A \cap B)

P(AB)=P((AB))=1P(AB) (De Morgan’s Law)P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) \text{ (De Morgan's Law)}

💡Examples

Problem 1:

A die is thrown once. Let event AA be 'getting a prime number' and event BB be 'getting an odd number'. Find the probability of P(AB)P(A \cup B).

Solution:

  1. Sample Space S={1,2,3,4,5,6}S = \{1, 2, 3, 4, 5, 6\}, so n(S)=6n(S) = 6.
  2. Event AA (prime numbers) = {2,3,5}\{2, 3, 5\}, so n(A)=3n(A) = 3 and P(A)=36=0.5P(A) = \frac{3}{6} = 0.5.
  3. Event BB (odd numbers) = {1,3,5}\{1, 3, 5\}, so n(B)=3n(B) = 3 and P(B)=36=0.5P(B) = \frac{3}{6} = 0.5.
  4. ABA \cap B (odd prime numbers) = {3,5}\{3, 5\}, so n(AB)=2n(A \cap B) = 2 and P(AB)=26=13P(A \cap B) = \frac{2}{6} = \frac{1}{3}.
  5. Apply addition theorem: P(AB)=P(A)+P(B)P(AB)=36+3626=46=23P(A \cup B) = P(A) + P(B) - P(A \cap B) = \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3}.

Explanation:

Identify the sample space, define the sets for each event, find their intersection, and use the Addition Theorem of probability to find the union.

Problem 2:

If P(A)=0.54P(A) = 0.54, P(B)=0.69P(B) = 0.69, and P(AB)=0.35P(A \cap B) = 0.35, calculate P(AB)P(A' \cap B').

Solution:

  1. First, find P(AB)P(A \cup B) using the formula: P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B).
  2. P(AB)=0.54+0.690.35=1.230.35=0.88P(A \cup B) = 0.54 + 0.69 - 0.35 = 1.23 - 0.35 = 0.88.
  3. By De Morgan's Law, AB=(AB)A' \cap B' = (A \cup B)'.
  4. P(AB)=P((AB))=1P(AB)P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B).
  5. P(AB)=10.88=0.12P(A' \cap B') = 1 - 0.88 = 0.12.

Explanation:

The probability of 'neither A nor B' (ABA' \cap B') is the complement of the probability of 'at least one of A or B' (ABA \cup B).