Probability - Other rules of Probability using Axiomatic Definition

Slide 1:

  • Today’s topic: Other rules of probability using the Axiomatic Definition
  • We will discuss three important rules: complement rule, addition rule, and multiplication rule
  • These rules help us compute probabilities in different scenarios

Slide 2:

  • Complement Rule:
    • The probability of an event A not occurring is denoted as P(A')
    • P(A’) = 1 - P(A)
    • Example: If the probability of getting a head in a fair coin toss is P(H) = 0.5, then the probability of not getting a head is P(H’) = 1 - 0.5 = 0.5

Slide 3:

  • Addition Rule:
    • For two mutually exclusive events A and B, the probability of A or B occurring is given by P(A or B) = P(A) + P(B)
    • Example: Let A be the event of getting an even number when rolling a fair six-sided die, and B be the event of getting an odd number. P(A) = 3/6 (as there are 3 even numbers), P(B) = 3/6 (as there are 3 odd numbers). Therefore, P(A or B) = 3/6 + 3/6 = 6/6 = 1.

Slide 4:

  • Addition Rule (contd.):
    • For two non-mutually exclusive events A and B, the probability of A or B occurring is given by P(A or B) = P(A) + P(B) - P(A and B)
    • Example: Let A be the event of rolling an even number, and B be the event of rolling a number greater than 3. P(A) = 3/6, P(B) = 3/6, P(A and B) = 1/6 (as there is only one outcome that satisfies both conditions). Therefore, P(A or B) = 3/6 + 3/6 - 1/6 = 5/6.

Slide 5:

  • Multiplication Rule:
    • For two independent events A and B, the probability of A and B occurring is given by P(A and B) = P(A) * P(B)
    • Example: Let A be the event of getting a head, and B be the event of getting a tail in two independent coin tosses. P(A) = 0.5, P(B) = 0.5. Therefore, P(A and B) = 0.5 * 0.5 = 0.25.

Slide 6:

  • Multiplication Rule (contd.):
    • For two dependent events A and B, the probability of A and B occurring is given by P(A and B) = P(A) * P(B|A)
    • P(B|A) denotes the conditional probability of B given A has already occurred
    • Example: Let A be the event of drawing a red card from a standard deck of 52 playing cards without replacement, and B be the event of drawing a black card from the remaining cards after A has already occurred. P(A) = 26/52 = 1/2, P(B|A) = 26/51. Therefore, P(A and B) = (1/2) * (26/51) = 1/4.

Slide 7:

  • Summary:
    • Complement Rule: P(A’) = 1 - P(A)
    • Addition Rule (Mutually Exclusive): P(A or B) = P(A) + P(B)
    • Addition Rule (Non-Mutually Exclusive): P(A or B) = P(A) + P(B) - P(A and B)
    • Multiplication Rule (Independent): P(A and B) = P(A) * P(B)
    • Multiplication Rule (Dependent): P(A and B) = P(A) * P(B|A)

Slide 8:

  • Let’s solve some practice problems!

Slide 9:

  • Problem 1:
    • A bag contains 4 red balls and 6 blue balls. What is the probability of drawing a red ball or a blue ball?
    • Solution: P(Red or Blue) = P(Red) + P(Blue) = 4/10 + 6/10 = 10/10 = 1

Slide 10:

  • Problem 2:
    • Two cards are drawn from a standard deck of 52 playing cards without replacement. What is the probability of both cards being red?
    • Solution: P(Red and Red) = (26/52) * (25/51) = 25/102 Here are slides 11 to 20 in Markdown format:

Slide 11:

  • Problem 3:
    • A box contains 10 red balls, 8 blue balls, and 6 green balls. If a ball is drawn at random, what is the probability of drawing a blue or green ball?
    • Solution: P(Blue or Green) = P(Blue) + P(Green) = 8/24 + 6/24 = 14/24 = 7/12

Slide 12:

  • Problem 4:
    • Two dice are rolled. What is the probability of getting a sum of 7 or 11?
    • Solution: P(Sum of 7 or 11) = P(Sum of 7) + P(Sum of 11) = 6/36 + 2/36 = 8/36 = 2/9

Slide 13:

  • Problem 5:
    • A card is drawn from a well-shuffled deck of 52 playing cards. What is the probability of drawing a king or a queen?
    • Solution: P(King or Queen) = P(King) + P(Queen) = 4/52 + 4/52 = 8/52 = 2/13

Slide 14:

  • Problem 6:
    • A jar contains 5 red marbles and 3 blue marbles. Two marbles are drawn without replacement. What is the probability of drawing a red marble on the first draw and a blue marble on the second draw?
    • Solution: P(Red and Blue) = P(Red) * P(Blue|Red) = (5/8) * (3/7) = 15/56

Slide 15:

  • Problem 7:
    • A pack of 52 playing cards is randomly shuffled. What is the probability of drawing a spade, a heart, and a diamond in that order?
    • Solution: P(Spade and Heart and Diamond) = P(Spade) * P(Heart|Spade) * P(Diamond|Spade and Heart) = (13/52) * (13/51) * (13/50) = 1/100

Slide 16:

  • Problem 8:
    • A bag contains 8 red balls, 4 blue balls, and 2 green balls. Three balls are drawn at random. What is the probability of drawing exactly 2 red balls?
    • Solution: P(2 Red balls) = P(2 Red and 1 Non-Red) = P(Red) * P(Red) * P(Non-Red) = (8/14) * (7/13) * (6/12) = 14/65

Slide 17:

  • Problem 9:
    • A committee of 4 members needs to be formed from a group of 8 men and 6 women. What is the probability that the committee consists of 2 men and 2 women?
    • Solution: P(2 Men and 2 Women) = P(2 Men) * P(2 Women) = (C(8, 2)/C(14, 4)) * (C(6, 2)/C(14, 4)) = (28/91) * (15/91) = 420/8281

Slide 18:

  • Problem 10:
    • A fair coin is tossed 5 times. What is the probability of getting exactly 3 heads?
    • Solution: P(3 Heads) = C(5, 3) * (1/2)^3 * (1/2)^2 = 10/32 = 5/16

Slide 19:

  • Summary:
    • Complement Rule: P(A’) = 1 - P(A)
    • Addition Rule (Mutually Exclusive): P(A or B) = P(A) + P(B)
    • Addition Rule (Non-Mutually Exclusive): P(A or B) = P(A) + P(B) - P(A and B)
    • Multiplication Rule (Independent): P(A and B) = P(A) * P(B)
    • Multiplication Rule (Dependent): P(A and B) = P(A) * P(B|A)

Slide 20:

  • Summary (contd.):
    • The complement rule helps find the probability of an event not occurring
    • The addition rule is used for finding the probability of events occurring together or individually
    • The multiplication rule is used for finding the probability of independent or dependent events occurring together
  1. Examples of Complement Rule:
  • Example 1: Consider a bag containing 4 red balls and 6 blue balls. Find the probability of not drawing a red ball.
  • Solution: P(Not Red) = 1 - P(Red) = 1 - 4/10 = 6/10 = 3/5
  • Example 2: A box contains 5 black socks and 7 white socks. What is the probability of not selecting a black sock?
  • Solution: P(Not Black) = 1 - P(Black) = 1 - 5/12 = 7/12
  1. Examples of Addition Rule (Mutually Exclusive):
  • Example 1: A standard deck of cards is shuffled. What is the probability of drawing either a red card or a face card?
  • Solution: P(Red or Face card) = P(Red) + P(Face card) = 26/52 + 12/52 = 38/52 = 19/26
  • Example 2: In a group of students, 16 are studying Physics and 20 are studying Chemistry. What is the probability of a student studying either Physics or Chemistry?
  • Solution: P(Physics or Chemistry) = P(Physics) + P(Chemistry) = 16/36 + 20/36 = 36/36 = 1
  1. Examples of Addition Rule (Non-Mutually Exclusive):
  • Example 1: A bag contains 3 red and 4 green balls. What is the probability of drawing either a red or a green ball?
  • Solution: P(Red or Green) = P(Red) + P(Green) - P(Red and Green) = 3/7 + 4/7 - 0 = 7/7 = 1
  • Example 2: A survey shows that 60% of people prefer tea, and 40% prefer coffee. What is the probability of a person preferring either tea or coffee?
  • Solution: P(Tea or Coffee) = P(Tea) + P(Coffee) - P(Tea and Coffee) = 60/100 + 40/100 - 0 = 1
  1. Examples of Multiplication Rule (Independent):
  • Example 1: A fair six-sided die is rolled twice. What is the probability of getting a 4 on the first roll and a 6 on the second roll?
  • Solution: P(4 and 6) = P(4) * P(6) = 1/6 * 1/6 = 1/36
  • Example 2: Two coins are tossed simultaneously. What is the probability of getting a head on the first coin and a tail on the second coin?
  • Solution: P(Head and Tail) = P(Head) * P(Tail) = 1/2 * 1/2 = 1/4
  1. Examples of Multiplication Rule (Dependent):
  • Example 1: A box contains 5 black balls and 3 white balls. Two balls are drawn without replacement. What is the probability of drawing a black ball followed by a white ball?
  • Solution: P(Black and White) = P(Black) * P(White|Black) = 5/8 * 3/7 = 15/56
  • Example 2: A bag contains 7 red marbles and 5 blue marbles. Two marbles are drawn without replacement. Find the probability of drawing a red marble first and a blue marble second.
  • Solution: P(Red and Blue) = P(Red) * P(Blue|Red) = 7/12 * 5/11 = 35/132
  1. Summary:
  • Complement Rule: P(A’) = 1 - P(A)
  • Addition Rule (Mutually Exclusive): P(A or B) = P(A) + P(B)
  • Addition Rule (Non-Mutually Exclusive): P(A or B) = P(A) + P(B) - P(A and B)
  • Multiplication Rule (Independent): P(A and B) = P(A) * P(B)
  • Multiplication Rule (Dependent): P(A and B) = P(A) * P(B|A)
  1. Additional Rule: Product Rule (Chain Rule)
  • The product rule is used to find the probability of the intersection of multiple events.
  • P(A and B and C) = P(A) * P(B|A) * P(C|A and B)
  • Example: If A, B, and C are independent events, the formula simplifies to P(A and B and C) = P(A) * P(B) * P(C)
  1. Additional Rule: Conditional Probability
  • Conditional probability represents the probability of an event occurring given that another event has already occurred.
  • P(A|B) denotes the probability of event A occurring given that event B has occurred.
  • Example: If the probability of event A is P(A) = 0.5 and the probability of event B is P(B) = 0.4, then the conditional probability P(A|B) depends on the relationship between A and B.
  1. Application in Real Life
  • Probability concepts find applications in various fields like finance, insurance, weather forecasting, sports analysis, etc.
  • Understanding these rules help in making informed decisions or predictions in uncertain situations.
  • Examples: Estimating stock market returns, predicting weather patterns, assessing the risk of an insurance claim, etc.
  1. Conclusion
  • Probability is a fundamental concept in mathematics with various applications in real-life situations.
  • The complement, addition, and multiplication rules provide a framework for calculating probabilities of events.
  • These rules help us make informed decisions and understand the likelihood of different outcomes.
  • Practice and application of these rules can enhance our understanding of probability and its significance in different domains.