91Ó°ÊÓ

Bayes' Theorem is a powerful tool in probability theory that allows us to update the probability of a hypothesis based on new evidence. It is essential when dealing with conditional probabilities. The theorem can be expressed as:

\[ P(A|B) = \frac{P(B|A)P(A)}{P(B)} \].

In this context, we wanted to find the probability that the first airline was responsible for the delay, given that the luggage didn't arrive on time. Let's break it down:

• Step 1: Identify the prior probability: Here, the prior probability is the probability of the first airline losing the luggage, \( P(D_1) = 0.4 \).
  
• Step 2: Identify the likelihood: This is the probability of the luggage being delayed if the first airline was responsible, which is also \( 0.4 \).
  
• Step 3: Calculate the total probability of the luggage not arriving on time, \( P(D) = 0.568 \).
  
• Step 4: Apply Bayes' Theorem: Plug these into the formula to get, \( P(D_1|D) = \frac{P(D_1)P(D)}{P(D)} = \frac{0.4}{0.568} \approx 0.704 \).
  

Bayes' Theorem helps us update our belief about the cause of the delay, given that a delay occurred.

Unabhängige Ereignisse (independent events) are crucial in understanding complex probability problems. Events are independent if the occurrence of one does not affect the occurrence of another. Mathematically, two events A and B are independent if:

\[ P(A \cap B) = P(A)P(B) \]

In the exercise, each luggage transfer (first, second, and third) happens independently. This means:

• First Transfer: The probability that the luggage was delayed by the first airline is \( P(D_1) = 0.4 \).
  
• Second Transfer: Similarly, \( P(D_2) = 0.2 \) does not depend on \( D_1 \).
  
• Third Transfer: \( P(D_3) = 0.1 \) is independent of both \( D_1 \) and \( D_2 \).
  

These independent probabilities help us find the total probability of the luggage not arriving on time. Total probability integrates the individual delays:
\[ P(D) = P(D_1) + P(D_2) + P(D_3) - P(D_1)P(D_2) - P(D_2)P(D_3) - P(D_1)P(D_3) + P(D_1)P(D_2)P(D_3) \].

Understanding independence ensures that we correctly account for multiple events impacting the overall outcome.

Wahrscheinlichkeitsverteilung (probability distribution) describes how the probabilities are assigned across different outcomes in a scenario. For discrete events, it involves listing each outcome and its associated probability. In our exercise:

• First Transfer: The probability of a delay is \( 0.4 \) and no delay is \( 0.6 \).
  
• Second Transfer: The delay probability is \( 0.2 \) and no delay is \( 0.8 \).
  
• Third Transfer: Delayed probability is \( 0.1 \) and no delay is \( 0.9 \).
  

The combination of these events forms the overall probability distribution, capturing every possible outcome of luggage arrival times. The probability distribution for total delay integrates independent probabilities:

• First: \( \ P(D_1) = 0.4 \)
  
• Second: \( P(D_2) = 0.2 \)
  
• Third: \( P(D_3) = 0.1 \)
  

The total delay's probability distribution follows:
\[ \ P(D) = 0.4 + 0.2 + 0.1 - 0.4 \times 0.2 - 0.2 \times 0.1 - 0.4 \times 0.1 + 0.4 \times 0.2 \times 0.1 = 0.568 \].

This reflects all combinations of delays across transfer stages, providing a complete picture of potential delays.