Slide 1 - Probability: Basic Concepts

  • Definition of probability
  • Sample space
  • Events
  • Probability of an event
  • Notation: P(A), where A is an event

Slide 2 - Types of Probability

  • Theoretical probability
  • Experimental probability
  • Relative frequency probability
  • Subjective probability
  • Probability Range: 0 ≤ P(A) ≤ 1

Slide 3 - Calculating Probability

  • Counting methods: factorial (n!)
  • Permutations: nPr
    • Formula: nPr = n! / (n-r)!
    • Example: Finding the number of ways to arrange letters in a word
  • Combinations: nCr
    • Formula: nCr = n! / r!(n-r)!
    • Example: Selecting a committee from a group of people

Slide 4 - Probability Rules

  • Union of events
    • P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
    • Example: Tossing a coin and rolling a die
  • Intersection of events
    • P(A ∩ B) = P(A) * P(B|A)
    • Example: Drawing cards from a deck with and without replacement

Slide 5 - Conditional Probability

  • Definition of conditional probability
  • Notation: P(B|A)
  • Calculation: P(B|A) = P(A ∩ B) / P(A)
  • Example: Probability of getting a red card given that it is a diamond

Slide 6 - Independent and Dependent Events

  • Independent events
    • Definition: P(B|A) = P(B)
    • Example: Tossing a coin and rolling a die
  • Dependent events
    • Definition: P(B|A) ≠ P(B)
    • Example: Drawing cards from a deck with replacement

Slide 7 - Mutually Exclusive Events

  • Definition of mutually exclusive events
  • Two events cannot occur at the same time
  • P(A ∩ B) = 0
  • Example: Rolling a die and getting an odd number or an even number

Slide 8 - Addition Rule for Mutually Exclusive Events

  • If A and B are mutually exclusive,
  • P(A ∪ B) = P(A) + P(B)
  • Example: Flipping a coin and rolling a die

Slide 9 - Addition Rule for Non-Mutually Exclusive Events

  • If A and B are not mutually exclusive,
  • P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
  • Example: Drawing a card from a deck and it being a heart or an ace

Slide 10 - Complementary Events

  • Definition of complementary events
  • P(A’) = 1 - P(A)
  • Example: Flipping a coin and getting heads or tails

Slide 11 - Probability - Problem on Nature of Outcome

  • Nature of outcome
    • Certain event: P(A) = 1, for an event A that is sure to happen
      • Example: Tossing a fair coin and getting heads or tails
    • Impossible event: P(A) = 0, for an event A that cannot happen
      • Example: Rolling a die and getting a 7
  • Probability as a number between 0 and 1
    • P(A) = 0.5 means there is a 50% chance of event A happening
    • P(A) = 0.8 means there is an 80% chance of event A happening
  • Equally likely outcomes
    • P(A) = number of favorable outcomes / total number of outcomes
    • Example: Rolling a fair 6-sided die

Slide 12 - Problem on Probability Range

  • Probability range: 0 ≤ P(A) ≤ 1
  • P(A) = 0, means event A is impossible
  • P(A) = 1, means event A is certain
  • Example 1: Find the probability of getting a prime number when rolling a fair 6-sided die.
    • Total number of outcomes = 6
    • Number of favorable outcomes = {2, 3, 5}
    • P(A) = 3/6 = 0.5
  • Example 2: Find the probability of selecting an even number from a deck of 52 playing cards.
    • Total number of outcomes = 52
    • Number of favorable outcomes = {2, 4, 6, 8, 10, 12, 14, …, 48, 50}
    • P(A) = 26/52 = 0.5

Slide 13 - Probability - Problem on Sample Space

  • Sample space: Set of all possible outcomes in an experiment
  • Example: Tossing a fair coin
    • Sample space: {H, T}
  • Problem: Find the sample space when rolling two fair dice.
    • Sample space = {(1,1), (1,2), …, (6,6)}

Slide 14 - Problem on Calculating Probability

  • Problem: Find the probability of selecting a odd number when rolling a fair 6-sided die.
    • Total number of outcomes = 6
    • Number of favorable outcomes = {1, 3, 5}
    • P(A) = 3/6 = 0.5
  • Problem: Find the probability of selecting a picture card (jack, queen, or king) from a deck of 52 playing cards.
    • Total number of outcomes = 52
    • Number of favorable outcomes = {J, Q, K} for all suits
    • P(A) = 12/52 = 3/13 ≈ 0.231

Slide 15 - Problem on Counting Methods: Factorial

  • Problem: How many different ways can the letters of the word “MATHS” be arranged?
    • Total number of letters = 5
    • Number of arrangements = 5! = 5 * 4 * 3 * 2 * 1 = 120

Slide 16 - Problem on Permutations

  • Problem: In how many ways can 3 different books be arranged on a bookshelf?
    • Total number of books = 3
    • Number of arrangements = 3P3 = 3! = 3 * 2 * 1 = 6
  • Problem: How many ways can 4 students be seated in a row?
    • Total number of students = 4
    • Number of arrangements = 4P4 = 4! = 4 * 3 * 2 * 1 = 24

Slide 17 - Problem on Combinations

  • Problem: How many ways can a committee of 3 students be selected from a group of 10 students?
    • Total number of students = 10
    • Number of students to be selected = 3
    • Number of combinations = 10C3 = 10! / (3! * (10-3)!) = 120 / (6 * 7) = 120 / 42 = 20

Slide 18 - Problem on Union of Events

  • Problem: Two fair dice are rolled. Find the probability of getting an even number or a number greater than 4.
    • P(Even) = 3/6 = 1/2
    • P(Greater than 4) = 2/6 = 1/3
    • P(Even ∪ Greater than 4) = P(Even) + P(Greater than 4) - P(Even ∩ Greater than 4) = 1/2 + 1/3 - 1/6 = 5/6

Slide 19 - Problem on Intersection of Events

  • Problem: A card is drawn from a well-shuffled deck of 52 playing cards. Find the probability of drawing a heart or a queen.
    • P(Heart) = 13/52 = 1/4
    • P(Queen) = 4/52 = 1/13
    • P(Heart ∩ Queen) = P(Queen) = 1/13
    • P(Heart ∩ Queen) = P(Queen) * P(Heart|Queen) = (1/13) * (1/52) = 1/676

Slide 20 - Problem on Conditional Probability

  • Problem: Two cards are drawn from a well-shuffled deck of 52 playing cards without replacement. What is the probability that the first card drawn is a king, given that the second card drawn is a queen?
    • P(King) = 4/52 = 1/13
    • P(Queen) = 4/51 (since one queen is already drawn)
    • P(King|Queen) = P(King ∩ Queen) / P(Queen) = (4/52) / (4/51) = (4/52) * (51/4) = 1/13 * 51/4 = 17/208 Here are slides 21 to 30 in markdown format:

Slide 21 - Probability - Problem on Nature of Outcome

  • Nature of outcome
    • Certain event: P(A) = 1, for an event A that is sure to happen
    • Impossible event: P(A) = 0, for an event A that cannot happen
  • Probability as a number between 0 and 1
    • P(A) = 0.5 means there is a 50% chance of event A happening
    • P(A) = 0.8 means there is an 80% chance of event A happening
  • Equally likely outcomes
    • P(A) = number of favorable outcomes / total number of outcomes
  • Example 1: Find the probability of getting a prime number when rolling a fair 6-sided die.
    • Total number of outcomes = 6
    • Number of favorable outcomes = {2, 3, 5}
    • P(A) = 3/6 = 0.5
  • Example 2: Find the probability of selecting an even number from a deck of 52 playing cards.
    • Total number of outcomes = 52
    • Number of favorable outcomes = {2, 4, 6, 8, 10, 12, 14, …, 48, 50}
    • P(A) = 26/52 = 0.5

Slide 22 - Problem on Probability Range

  • Probability range: 0 ≤ P(A) ≤ 1
  • P(A) = 0, means event A is impossible
  • P(A) = 1, means event A is certain
  • Example 1: Find the probability of getting a tail when flipping a fair coin.
    • P(Tail) = 1/2 = 0.5
  • Example 2: Find the probability of selecting a red card from a deck of 52 playing cards.
    • P(Red) = 26/52 = 0.5

Slide 23 - Probability - Problem on Sample Space

  • Sample space: Set of all possible outcomes in an experiment
  • Example: Tossing a fair coin
    • Sample space: {Head, Tail}
  • Problem: Find the sample space when rolling two fair dice.
    • Sample space = {(1,1), (1,2), …, (6,6)}

Slide 24 - Problem on Calculating Probability

  • Problem: Find the probability of getting an even number when rolling a fair 6-sided die.
    • Total number of outcomes = 6
    • Number of favorable outcomes = {2, 4, 6}
    • P(Even) = 3/6 = 0.5
  • Problem: Find the probability of selecting a King from a deck of 52 playing cards.
    • Total number of outcomes = 52
    • Number of favorable outcomes = {King of Hearts, King of Diamonds, King of Spades, King of Clubs}
    • P(King) = 4/52 = 1/13 ≈ 0.077

Slide 25 - Problem on Counting Methods: Factorial

  • Problem: How many different ways can the letters of the word “STATISTICS” be arranged?
    • Total number of letters = 10 (2 “S”, 3 “T”, 2 “I”, 1 “A”, 1 “C”)
    • Number of arrangements = 10! / (2! * 3! * 2! * 1! * 1!) ≈ 120,960

Slide 26 - Problem on Permutations

  • Problem: In how many ways can 5 books be arranged on a bookshelf?
    • Total number of books = 5
    • Number of arrangements = 5! = 5 * 4 * 3 * 2 * 1 = 120
  • Problem: How many ways can a group of 3 students be selected from a class of 10 students to form a committee?
    • Total number of students = 10
    • Number of students to be selected = 3
    • Number of arrangements = 10P3 = 10! / (10-3)! = 10! / 7! = 10 * 9 * 8 = 720

Slide 27 - Problem on Combinations

  • Problem: How many ways can a committee of 4 students be selected from a group of 10 students?
    • Total number of students = 10
    • Number of students to be selected = 4
    • Number of combinations = 10C4 = 10! / (4! * (10-4)!) = 10! / (4! * 6!) ≈ 210

Slide 28 - Problem on Union of Events

  • Problem: Two cards are drawn from a well-shuffled deck of 52 playing cards. Find the probability of getting a heart or a diamond.
    • P(Heart) = 13/52 = 1/4
    • P(Diamond) = 13/52 = 1/4
    • P(Heart ∪ Diamond) = P(Heart) + P(Diamond) = 1/4 + 1/4 = 1/2

Slide 29 - Problem on Intersection of Events

  • Problem: A card is drawn from a well-shuffled deck of 52 playing cards. Find the probability of getting a spade and a face card.
    • P(Spade) = 13/52 = 1/4
    • P(Face Card) = 12/52 = 3/13
    • P(Spade ∩ Face Card) = P(Spade) * P(Face Card|Spade) = (1/4) * (3/12) = 1/16

Slide 30 - Problem on Conditional Probability

  • Problem: A jar contains 12 red balls and 8 blue balls. Two balls are drawn without replacement. Find the probability that the first ball drawn is red, given that the second ball drawn is blue.
    • P(Red) = 12/(12+8) = 12/20 = 3/5
    • P(Blue) = 8/(12+8) = 8/20 = 2/5
    • P(Red|Blue) = P(Red ∩ Blue) / P(Blue) = 12/20 * 8/19 / (2/5) = 96/380 = 24/95