2.4 Conditional Probability

2.4 Conditional Probability

Learning Objectives

  • Upon completion of this section, you should be able to:
    • Compute conditional probabilities.
    • Compute probability using the multiplication rule.

Definition of Conditional Probability

  • Conditional Probability: The conditional probability of event A given event B is the probability of A occurring provided that B has already occurred.
    • This is mathematically denoted as ( P(A|B) ), which translates to "the probability of A given B."
    • Interpretation: It measures how the probability of A changes when additional information (event B) is available.

Real-World Applications

  • Updating probabilities based on new information is essential for understanding scenarios in various fields, such as:
    • Weather forecasting.
    • Medical diagnoses.

Example - Rolling a Die

  • Scenario: You have a fair six-sided die:

    1. The initial probability of rolling a five is ( P(F) = \frac{1}{6} ).
    2. If you know the number rolled is odd, the possible outcomes are 1, 3, or 5. Hence, the conditional probability becomes:

    • Given that the roll was odd, the outcomes are reduced, and we recalculate:

    • New probability of rolling a five given it's odd, ( P(F|O) = \frac{P(F \cap O)}{P(O)} ). Matshematically:

      [ P(F|O) = \frac{1/6}{3/6} = \frac{1}{3} ]

      • Thus, the new answer reflecting the additional information of oddness is ( P(F|O) = \frac{1}{3} ).
      • Furthermore, knowledge of event B (being odd) affects the likelihood of event A (rolling a five).

General Formula

  • The General Formula for conditional probability is: [ P(A|B) = \frac{P(A \cap B)}{P(B)} ]
    • Where:
    • ( P(A \cap B) ) is the probability that both events A and B occur.
    • ( P(B) ) is the probability that event B occurs.

Examples and Solutions

Example 1 - Probability of Five Given Odd
  • Query: Find the probability that the number rolled is a five, given that it is odd.
    • Events defined as:
    • ( F = {5} ): Event that a five is rolled.
    • ( O = {1, 3, 5} ): Event that an odd number is rolled.
    • Solution Calculation:
    • Sample space ( S = {1, 2, 3, 4, 5, 6} ).
    • Using formula:
    • [ P(F|O) = \frac{P(F \cap O)}{P(O)} ] with:
      • ( F \cap O = {5} \Rightarrow P(F \cap O) = \frac{1}{6} )
      • ( O = {1, 3, 5} \Rightarrow P(O) = \frac{3}{6} )
    • Thus:
      [ P(F|O) = \frac{1/6}{3/6} = \frac{1}{3} ]
    • Reversed Query: Find the probability that the number rolled is odd, given that it is a five:
    • Using the same logic:
    • [ P(O|F) = \frac{P(O \cap F)}{P(F)} = 1 ] since rolling a five guarantees that the outcome is odd.

Two-Way Frequency Table Example

Example 2 - Gender and Age at First Marriage

  • Data Setup: A sample of 902 married individuals under 40 classified by gender and age at first marriage is shown in a two-way table:

    • | Age Group (at first marriage) | Male | Female | Total |
    • |-------------------------------|------|--------|-------|
    • | Teenagers | 43 | 82 | 125 |
    • | Twenties | 299 | 592 | 891 |
    • | Thirties | 114 | 185 | 299 |
    • | Total | 450 | 452 | 902 |

  • Calculating Probabilities:

    • Probability of being a teenager at first marriage:
    • [ P(E) = \frac{125}{902} \approx 0.139 ]
    • Probability of being in their teens, given male:
    • [ P(E|M) = \frac{43}{450} \approx 0.096 ]
Example 3 - Hypertension and Weight
  • Probabilities:
    • Proportion of people overweight & hypertensive: 0.09.
    • Proportion not overweight but hypertensive: 0.11.
    • Compare the probabilities for better understanding of the relationship between weight and hypertension:
    • [ P(H|O) = \frac{0.09}{0.09 + 0.02} \approx 0.8182 ]
    • [ P(H|O^C) = \frac{0.11}{0.11 + 0.78} \approx 0.1236 ]
Conditional Probability in Contingency Tables
  • The values presented allow calculation without needing separate probability computations.

Multiplication Rule

  • General Multiplication Rule: Describes the probability of events A and B occurring together, expressed as:
    [ P(A \cap B) = P(B|A) \cdot P(A) = P(A|B) \cdot P(B) ]
  • This highlights the interrelation between the events and their conditional probabilities.
  • Example: If you are drawing cards from a deck, calculating probabilities for two draws in sequence can utilize these principles effectively.

Important Definitions

  • Independent Events: Two events A and B are defined as independent if: [ P(A|B) = P(A) ] (the occurrence of B does not affect the probability of A)
    • Example context: Storm damage does not affect the likelihood of a prior burglary.
  • Dependent Events: If the above condition does not hold, then events A and B are dependent. Understanding this distinction is crucial for proper probability calculations.

Summary

  • Conditional probabilities provide insights into events affected by prior occurrences.
  • Mastery of the multiplication rule and the distinction of dependency among events are fundamental in probability theory.