lesson 12multiplicatiom rule amd conditional probability

Independence of Events

  • Definition of Independence:
    • Two events, A and B, are independent if the occurrence of one does not affect the probability of the other.
    • Mathematically, if P(A) is the probability of event A occurring, and P(B|A) is the probability of event B occurring given A has occurred, then:
      P(B∣A)=P(B)P(B|A) = P(B)
    • If this condition holds true, then events A and B are independent.

Multiplication Rule for Independent Events

  • Multiplication Rule:
    • The probability that both events A and B occur (denoted as P(A and B)) when they are independent is given by:
      P(A and B)=P(A)×P(B)P(A \text{ and } B) = P(A) \times P(B)
    • Important to remember:
    • Use "and" for multiplication rule (P(A and B)).
    • Use "or" for addition rule (P(A or B)).

Example 1: Smartphones

  • Problem: In a study, 87% of students have a smartphone.

    • Find the probability that both of two randomly selected students have a smartphone.
    • Let
      P(A)=P(B)=0.87P(A) = P(B) = 0.87
    • Since they are independent, apply the multiplication rule:
      P(A and B)=0.87×0.87=0.7569P(A \text{ and } B) = 0.87 \times 0.87 = 0.7569

  • Part B: Find the probability that the first student has a smartphone and the second does not:

    • Use the complement for the second student, where P(B′)=1−P(B)=1−0.87=0.13P(B') = 1 - P(B) = 1 - 0.87 = 0.13
    • Calculate:
      P(A and B′)=P(A)×P(B′)=0.87×0.13=0.1131P(A \text{ and } B') = P(A) \times P(B') = 0.87 \times 0.13 = 0.1131

  • Part C: Find the probability that both students do not have a smartphone:

    • Calculate as follows:
      P(A′)=0.13(complement)P(A') = 0.13 \quad (complement)
      P(A′ and B′)=0.13×0.13=0.0169P(A' \text{ and } B') = 0.13 \times 0.13 = 0.0169

Example 2: Loaded Coin

  • Problem: A loaded coin shows heads 72% of the time. Toss the coin twice and find:
    1. Both heads:
    • P(H and H)=0.72×0.72=0.5184P(H \text{ and } H) = 0.72 \times 0.72 = 0.5184
    1. Heads then tails:
    • P(H and T)=0.72×0.28=0.2016P(H \text{ and } T) = 0.72 \times 0.28 = 0.2016
    1. Tails then heads:
    • P(T and H)=0.28×0.72=0.2016P(T \text{ and } H) = 0.28 \times 0.72 = 0.2016
    1. Both tails:
    • P(T and T)=0.28×0.28=0.0784P(T \text{ and } T) = 0.28 \times 0.28 = 0.0784

Complement Rule and At Least One

  • Finding the Probability of At Least One Event:
    • To find the probability of at least one occurrence, calculate the complement:
    • For three children, the probability of no boys is:
      P(0 boys)=P(G)×P(G)×P(G)=0.3×0.3×0.3=0.027P(0 \text{ boys}) = P(G) \times P(G) \times P(G) = 0.3 \times 0.3 \times 0.3 = 0.027
    • Therefore, the probability of at least one boy is:
      P(at least 1 boy)=1−P(0 boys)=1−0.027=0.973P(\text{at least 1 boy}) = 1 - P(0 \text{ boys}) = 1 - 0.027 = 0.973

Example with Multiple Choice Quiz

  • Problem: For a five-question multiple choice quiz (one correct answer out of four choices), find the probability of guessing at least one correctly:
    1. Calculate probability of getting zero correct answers (all guessing wrong):
      • Each question has a 75% chance of being answered incorrectly.
      • P(0)=0.755P(0) = 0.75^5
    2. Find:
      P(at least 1 correct)=1−P(0)P(\text{at least 1 correct}) = 1 - P(0)