Dependent Events

Introduction

Understanding Conditional Probability is crucial in numerous real-world applications such as weather forecasting, medicine, and business. Conditional probability is defined as the likelihood of an event occurring given another event has already taken place. Consider a common scenario: choosing to carry an umbrella because it is cloudy and rain is predicted.

Importance

• It is essential for making informed predictions in various fields.
• Assists in decision-making processes by accounting for existing conditions.

Define Conditional Probability

Conditional Probability: The probability of an event occurring given another event has occurred.

Formula

The conditional probability is calculated using:

$P(A|B) = \frac{P(A \cap B)}{P(B)}$

• P(A|B): Denotes the probability of event A given event B has occurred.
• $P(A \cap B)$: The probability of both events A and B occurring simultaneously.
• $P(B)$: The probability of the occurrence of event B.

Observe the significance of this concept in everyday situations such as weather predictions or medical testing.

Venn Diagrams

Purpose: Venn diagrams play a crucial role in understanding relationships in probability. They simplify complex probability scenarios by clearly illustrating event overlaps and interactions.

Worked Examples

Scenario Example: Calculating the combined interest in mathematics and science among students.

Steps:

1. Identify what we're looking for: the probability of students interested in both subjects
2. Given information:
   • 60% of students like mathematics
   • 40% of students like science
   • 25% of students like both subjects
3. Apply the formula: $P(A \cap B) = P(A) \times P(B|A)$
4. Alternatively, we can directly use the overlap from our Venn diagram: P(Mathematics ∩ Science) = 0.25

Common Mistakes and Clarifications

Checklist:

• Correctly identify pairs of events.
• Accurately compute intersections and their probabilities.

Practice Problems

Examples with Solutions:

1. Traffic Problem:

   • Question: If the probability of rain is 0.3, the probability of traffic is 0.6, and the probability of both rain and traffic is 0.25, what is the probability of traffic given that it rains?
   • Solution: $P(\text{Traffic}|\text{Rain}) = \frac{P(\text{Traffic} \cap \text{Rain})}{P(\text{Rain})} = \frac{0.25}{0.3} = 0.833$ or approximately 83.3%

2. Games Problem:

   • Question: If the probability of practising is 0.7, the probability of winning is 0.4, and the probability of both practising and winning is 0.35, what is the probability of winning given that practice has taken place?
   • Solution: $P(\text{Win}|\text{Practice}) = \frac{P(\text{Win} \cap \text{Practice})}{P(\text{Practice})} = \frac{0.35}{0.7} = 0.5$ or 50%

3. Healthcare Problem:

   • Question: If a disease affects 2% of the population and a test for this disease has 95% accuracy (true positive) and 10% false positive rate, what is the probability that a person has the disease given they tested positive?
   • Solution:
     • $P(D) = 0.02$ (probability of disease)
     • $P(+|D) = 0.95$ (probability of positive test given disease)
     • $P(+|\text{not }D) = 0.1$ (probability of positive test given no disease)
     • Using Bayes' theorem: $P(D|+) = \frac{P(+|D) \times P(D)}{P(+)} = \frac{0.95 \times 0.02}{0.95 \times 0.02 + 0.1 \times 0.98} = \frac{0.019}{0.019 + 0.098} = \frac{0.019}{0.117} = 0.162$ or approximately 16.2%