Probability - Two fair dice tossing

  • Definition of Probability
  • Sample Space
  • Event
  • Complementary Event
  • Probability of an Event

Definition of Probability

  • Probability is a measure of the likelihood of an event occurring.
  • It is represented by a number between 0 and 1.
  • Probability of 0 means the event will not occur.
  • Probability of 1 means the event is certain to occur.

Sample Space

  • Sample Space is the set of all possible outcomes of an experiment.
  • In the case of tossing two fair dice, the sample space would be {1, 2, 3, 4, 5, 6} for each dice.
  • The overall sample space would be the set of all possible pairs of outcomes.

Event

  • An event is a subset of the sample space.
  • It represents a specific outcome or a combination of outcomes.
  • In the case of tossing two fair dice, an event could be “getting a sum of 7” or “getting a double”.

Complementary Event

  • The complementary event of an event A represents all outcomes that are not in A.
  • It is denoted by Ā or A'.
  • The probability of the complementary event is given by P(Ā) = 1 - P(A). Example: If A represents “getting a sum of 7”, then Ā represents “not getting a sum of 7”.

Probability of an Event

  • The probability of an event A is denoted by P(A).
  • It is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.
  • For equally likely outcomes, P(A) = Number of favorable outcomes / Total number of possible outcomes. Example: If A represents “getting a sum of 7” and there are 6 possible outcomes for getting a sum of 7, then P(A) = 6/36 = 1/6.

Example - Tossing Two Fair Dice

  • Let’s consider the event A = “getting a sum of 7”.
  • The favorable outcomes for A are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1).
  • The total number of possible outcomes for tossing two fair dice is 6 * 6 = 36.
  • Therefore, P(A) = 6/36 = 1/6.

Example - Tossing Two Fair Dice cont.

  • Let’s consider the event B = “getting a double”.
  • The favorable outcomes for B are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6).
  • The total number of possible outcomes for tossing two fair dice is 6 * 6 = 36.
  • Therefore, P(B) = 6/36 = 1/6.

Example - Tossing Two Fair Dice cont.

  • Let’s consider the event C = “getting a sum less than or equal to 5”.
  • The favorable outcomes for C are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), and (5, 1).
  • The total number of possible outcomes for tossing two fair dice is 6 * 6 = 36.
  • Therefore, P(C) = 15/36 = 5/12.

Example - Tossing Two Fair Dice cont.

  • Let’s consider the event D = “getting a sum greater than 9”.
  • The favorable outcomes for D are (4, 6), (5, 5), (5, 6), (6, 4), (6, 5), and (6, 6).
  • The total number of possible outcomes for tossing two fair dice is 6 * 6 = 36.
  • Therefore, P(D) = 6/36 = 1/6.

Example - Tossing Two Fair Dice cont.

  • Let’s consider the event E = “getting a sum of 2 or a sum of 12”.
  • The favorable outcomes for E are (1, 1) and (6, 6).
  • The total number of possible outcomes for tossing two fair dice is 6 * 6 = 36.
  • Therefore, P(E) = 2/36 = 1/18.

Example - Tossing Two Fair Dice cont.

  • Let’s consider the event F = “getting a sum of 7 or a double”.
  • The favorable outcomes for F are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6).
  • The total number of possible outcomes for tossing two fair dice is 6 * 6 = 36.
  • Therefore, P(F) = 12/36 = 1/3.

Example - Tossing Two Fair Dice cont.

  • Let’s consider the event G = “getting a sum of 10 or less than 4”.
  • The favorable outcomes for G are (1, 1), (2, 1), (1, 2), (1, 3), (3, 1), (2, 2), (2, 3), and (3, 2).
  • The total number of possible outcomes for tossing two fair dice is 6 * 6 = 36.
  • Therefore, P(G) = 8/36 = 2/9.

Conditional Probability

  • Conditional Probability is the probability of an event given that another event has occurred.
  • It is denoted by P(A | B), read as “the probability of A given B”.
  • The conditional probability of A given B is given by P(A | B) = P(A ∩ B) / P(B), where P(B) ≠ 0.

Example - Conditional Probability

  • Let’s consider the events A = “getting a sum of 7” and B = “getting a double”.
  • The favorable outcomes for A ∩ B are (1, 6), (6, 1).
  • The favorable outcomes for B are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6).
  • The probability of A given B is P(A | B) = P(A ∩ B) / P(B) = 2/6 = 1/3.

Independent Events

  • Events A and B are said to be independent if the occurrence of A does not affect the probability of B, and vice versa.
  • Mathematically, P(A ∩ B) = P(A) * P(B). Example: Tossing two fair coins is an example of independent events.

Dependent Events

  • Events A and B are said to be dependent if the occurrence of one event affects the probability of the other event.
  • Mathematically, P(A ∩ B) = P(A) * P(B | A) = P(B) * P(A | B), where P(A) and P(B) ≠ 0. Example: Drawing two cards from a deck without replacement is an example of dependent events.

Probability - Two fair dice tossing

  1. Conditional Probability - Example
  • Let’s consider the events A = “getting a sum of 7” and B = “getting a double”.
  • The favorable outcomes for A ∩ B are (1, 6), (6, 1).
  • The favorable outcomes for B are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6).
  • The probability of A given B is P(A | B) = P(A ∩ B) / P(B) = 2/6 = 1/3.
  1. Independent Events - Definition
  • Two events A and B are independent if the occurrence of A does not affect the probability of B, and vice versa.
  • Mathematically, P(A ∩ B) = P(A) * P(B).
  1. Independent Events - Example
  • Let’s consider the events A = “getting a sum less than or equal to 5” and B = “getting a double”.
  • The favorable outcomes for A ∩ B are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5).
  • The favorable outcomes for A are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (4, 1), (4, 2), and (5, 1).
  • The favorable outcomes for B are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), and (6, 6).
  • P(A ∩ B) = 5/36 = P(A) * P(B) = 15/36 * 6/36 = 5/36.
  • Since P(A ∩ B) = P(A) * P(B), events A and B are independent.
  1. Dependent Events - Definition
  • Two events A and B are dependent if the occurrence of one event affects the probability of the other event.
  • Mathematically, P(A ∩ B) = P(A) * P(B | A) = P(B) * P(A | B), where P(A) and P(B) ≠ 0.
  1. Dependent Events - Example
  • Let’s consider the events A = “drawing a red card from a deck” and B = “drawing a black card from the remaining deck without replacement”.
  • The favorable outcomes for A ∩ B are the same as the favorable outcomes for B because drawing a black card doesn’t affect the probability of drawing a red card.
  • The favorable outcomes for A are 26 (out of 52 cards), and the favorable outcomes for B are 26 (out of 51 cards).
  • P(A ∩ B) = 26/52 * 26/51 ≠ P(A) * P(B) = 26/52 * 26/52, so events A and B are dependent.
  1. Law of Total Probability
  • The Law of Total Probability states that for any event A and any partition of the sample space into mutually exclusive events B1, B2, …, Bn:
    • P(A) = P(A | B1) * P(B1) + P(A | B2) * P(B2) + … + P(A | Bn) * P(Bn).
  1. Law of Total Probability - Example
  • Let’s consider the events A = “getting a sum of 7” and B1, B2, …, B6 = “getting a double with each number on the first dice”.
  • The favorable outcomes for A | Bi are (i, 7-i) for i = 1 to 6.
  • The favorable outcomes for Bi are (i, i) for i = 1 to 6.
  • P(A) = P(A | B1) * P(B1) + P(A | B2) * P(B2) + … + P(A | B6) * P(B6) = (1/6) * (1/6) + (1/6) * (1/6) + … + (1/6) * (1/6) = 1/6.
  1. Bayes’ Theorem
  • Bayes’ Theorem allows us to update the probability of an event based on new information.
  • It is given by: P(A | B) = (P(B | A) * P(A)) / P(B), where P(A) and P(B) ≠ 0.
  1. Bayes’ Theorem - Example
  • Let’s consider the events A = “getting a sum of 7” and B = “getting a double”.
  • We already calculated P(A | B) = 1/3.
  • We also know that P(A) = 1/6 and P(B) = 1/6.
  • P(B | A) = P(A ∩ B) / P(A) = 2/36 / 6/36 = 2/6 = 1/3.
  • P(A | B) = (P(B | A) * P(A)) / P(B) = (1/3) * (1/6) / (1/6) = 1/3.
  1. Summary In this lecture, we covered:
  • Definition of Probability
  • Sample Space and Event
  • Complementary Event
  • Probability of an Event
  • Conditional Probability
  • Independent and Dependent Events
  • Law of Total Probability
  • Bayes’ Theorem These concepts are fundamental in understanding and solving probability problems and can be applied to various real-life scenarios.