krit.club logo

Statistics and Probability - Events and their Algebra

Grade 11ICSE

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

🔑Concepts

•

Random Experiment and Sample Space (SS): A random experiment is a process where the result is uncertain. The set of all possible outcomes is called the Sample Space (SS). Visually, this is represented by a large rectangular box (Universal Set) that contains every possible individual outcome as a point within it.

•

Events (Simple and Compound): An event is a subset of the sample space. A 'Simple Event' consists of only one outcome, while a 'Compound Event' consists of two or more. Visually, an event AA is depicted as a closed loop or circle inside the sample space rectangle.

•

Union of Events (A∪BA \cup B): The union represents the event that AA occurs, or BB occurs, or both occur (at least one occurs). Visually, this is the entire region covered by circles AA and BB combined. In terms of logic, it corresponds to the word 'OR'.

•

Intersection of Events (A∩BA \cap B): The intersection represents the event that both AA and BB occur simultaneously. Visually, this is the overlapping region where circles AA and BB cross each other. In terms of logic, it corresponds to the word 'AND'.

•

Complementary Events (A′A' or AcA^c): The complement of AA consists of all outcomes in the sample space SS that are not in AA. Visually, if circle AA is the area of interest, the complement is everything inside the rectangle that is outside circle AA. It satisfies P(A)+P(A′)=1P(A) + P(A') = 1.

•

Mutually Exclusive Events: Two events are mutually exclusive if they cannot happen at the same time, meaning A∩B=ϕA \cap B = \phi (an empty set). Visually, these are represented as two separate circles that do not overlap or touch each other.

•

Exhaustive Events: Events E1,E2,...,EnE_1, E_2, ..., E_n are exhaustive if their union covers the entire sample space (E1∪E2∪...∪En=SE_1 \cup E_2 \cup ... \cup E_n = S). Visually, if you combine all these event regions, they perfectly fill the entire sample space rectangle.

•

Difference of Events (A−BA - B): The event A−BA - B (or A∩B′A \cap B') represents outcomes that are in AA but not in BB. Visually, this is the portion of circle AA that does not overlap with circle BB, resembling a 'crescent moon' shape if they overlap.

📐Formulae

Definition of Probability: P(A)=n(A)n(S)P(A) = \frac{n(A)}{n(S)}

Range of Probability: 0≤P(A)≤10 \le P(A) \le 1

Complementary Rule: P(A′)=1−P(A)P(A') = 1 - P(A)

Addition Theorem (General): P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)

Addition Theorem (Mutually Exclusive): P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B) since P(A∩B)=0P(A \cap B) = 0

Probability of 'Only A': P(A−B)=P(A∩B′)=P(A)−P(A∩B)P(A - B) = P(A \cap B') = P(A) - P(A \cap B)

De Morgan's First Law: P(A′∩B′)=P((A∪B)′)=1−P(A∪B)P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B)

De Morgan's Second Law: P(A′∪B′)=P((A∩B)′)=1−P(A∩B)P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B)

💡Examples

Problem 1:

A fair die is rolled once. Let event AA be getting an even number and event BB be getting a number greater than 3. Find P(A∪B)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 (Even) = {2,4,6}\{2, 4, 6\}, so n(A)=3n(A) = 3 and P(A)=36P(A) = \frac{3}{6}.
  3. Event BB (>3) = {4,5,6}\{4, 5, 6\}, so n(B)=3n(B) = 3 and P(B)=36P(B) = \frac{3}{6}.
  4. Intersection A∩BA \cap B (Even AND >3) = {4,6}\{4, 6\}, so n(A∩B)=2n(A \cap B) = 2 and P(A∩B)=26P(A \cap B) = \frac{2}{6}.
  5. Apply Addition Theorem: P(A∪B)=P(A)+P(B)−P(A∩B)P(A \cup B) = P(A) + P(B) - P(A \cap B)
  6. P(A∪B)=36+36−26=46=23P(A \cup B) = \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6} = \frac{2}{3}.

Explanation:

We identify the individual probabilities and the overlapping outcomes (intersection) to avoid double-counting when calculating the union.

Problem 2:

Given P(A)=0.5P(A) = 0.5, P(B)=0.6P(B) = 0.6, and P(A∩B)=0.3P(A \cap B) = 0.3, find the probability that neither AA nor BB occurs.

Solution:

  1. The event 'Neither AA nor BB' is represented by (A′∩B′)(A' \cap B').
  2. By De Morgan's Law, P(A′∩B′)=P((A∪B)′)P(A' \cap B') = P((A \cup B)').
  3. First, find P(A∪B)=P(A)+P(B)−P(A∩B)=0.5+0.6−0.3=0.8P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.5 + 0.6 - 0.3 = 0.8.
  4. Now, find the complement: P((A∪B)′)=1−P(A∪B)=1−0.8=0.2P((A \cup B)') = 1 - P(A \cup B) = 1 - 0.8 = 0.2.

Explanation:

This problem uses the relationship between the union and its complement via De Morgan's Laws to find the probability of neither event occurring.