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tanszek:oktatas:techcomm:conditional_probability

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tanszek:oktatas:techcomm:conditional_probability [2024/09/30 10:49] – kneheztanszek:oktatas:techcomm:conditional_probability [2024/10/13 17:27] (current) – kissa
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 === Non-independent events === === Non-independent events ===
  
-How can we calculate the result in a case where two events are not independent? This means that if one event occurs, it will directly affect the probability of the other event.+How can we calculate the result when two events are not independent? If one event occurs, it will directly affect the probability of the other event.
  
 If events **A** and **B** are complex events that will not exclude each other, we have a so-called conditional probability (event A affects event B). If events **A** and **B** are complex events that will not exclude each other, we have a so-called conditional probability (event A affects event B).
  
-Notation: \(p(A | B) \)+**Notation**: \(p(A | B) \)
  
-In this case we mean the relative frequency which compares the sum of all probability to the probability of event B (probability of it's occurrance).+In this case, we mean the relative frequency, which compares the sum of all probabilities to the probability of event B (the probability of its occurrence).
  
-$$ p(A|B) = \frac{k_{AB}}{k_b} = \frac{\frac{k_{AB}}{k}}{\frac{k_{B}}{k_b}} = \frac{p(A \cap B)}{p(B)} $$+$$ p(A|B) = \frac{k_{AB}}{k_b} = \frac{\frac{k_{AB}}{k}}{\frac{k_{B}}{k}} = \frac{p(A \cap B)}{p(B)} $$
  
 So we can get to the conclusion: So we can get to the conclusion:
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 1.) \( p(A \cap B) \): 1.) \( p(A \cap B) \):
 This represents the probability that both events  This represents the probability that both events 
-A and B occur simultaneously. It is also known as the probability of the intersection of A and B.+A and B occur simultaneously, which is also known as the probability of their intersection.
  
 2.) \(  p(A|B) \): This is the conditional probability of event A occurring given that event B has already occurred. It tells us how likely A s to happen under the condition that B has happened. 2.) \(  p(A|B) \): This is the conditional probability of event A occurring given that event B has already occurred. It tells us how likely A s to happen under the condition that B has happened.
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 **What the Formula Says?** **What the Formula Says?**
  
-The formula states that the probability of both events A and B occurring together, is equal to the probability of B occurring multipliend by the probability of A occurring given that B has already occurred.+The formula states that the probability of events **A** and **B** occurring together, is equal to the probability of B occurring multiplied by the probability of A occurring given that B has already occurred.
  
-**Example**:+**Example 1**:
  
-A manufacturer needs to produce a shaft with two critical dimensions: length (L) and diameter (D). Tolerances of \( L \pm \Delta \) and \( D \pm \Delta \) is allowed. After inspecting 180 components, the results are as follows:+A manufacturer must produce a shaft with two critical dimensions: length (L) and diameter (D). Tolerances of \( L \pm \Delta \) and \( D \pm \Delta \) is allowed. After inspecting 180 components, the results are as follows:
  
  
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 | Both dimensions are faulty \( ( A \cap B ) \) | 4 | | Both dimensions are faulty \( ( A \cap B ) \) | 4 |
 | Only the length 𝐿 is faulty \((C)\) | 6 | | Only the length 𝐿 is faulty \((C)\) | 6 |
-| Only the diameter 𝐷 is faulty \(D)\) | 8 |+| Only the diameter 𝐷 is faulty \((D)\) | 8 |
  
 **Question 1**: What are the probabilities of events \(A\) and \(B\)? **Question 1**: What are the probabilities of events \(A\) and \(B\)?
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 $$ p(A \mid B) = \frac{\text{both dimensions are faulty}}{\text{diameter is faulty}} = \frac{4}{12} = 0.3333 $$ $$ p(A \mid B) = \frac{\text{both dimensions are faulty}}{\text{diameter is faulty}} = \frac{4}{12} = 0.3333 $$
  
-Since this does not match with the product \( p(A) p(B)\), we can conclude that the two events are **not independent**!+Since this does not match with the product \( p(A) \cdot p(B)\), we can conclude that the two events are **not independent**!
  
 Thus, the joint probability can also be calculated differently: Thus, the joint probability can also be calculated differently:
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 where \( E = A \cap B \) where \( E = A \cap B \)
 +
 +
 +**Example 2**:
 +Find the conditional probability of a machine breakdown given that preventive maintenance was performed.
 +
 +We have the following information:
 +  * The probability that the machine breaks down (Event A) is \( p(A) = 0.10 \)
 +  * The probability that the machine undergoes preventive maintenance (Event B) is \( p(B) = 0.30 \)
 +  * The probability that the machine both breaks down and has undergone preventive maintenance is \( p(A \cap B) = 0.015 \)
 +
 +To calculate the conditional probability of the machine breaking down given that it underwent preventive maintenance, we apply the conditional probability formula: 
 +
 +$$ p(A|B) = \frac{p(A \cap B)}{p(B)} $$
 +
 +Substitute the values:
 +
 +$$ p(A|B) = \frac{0.015}{0.30} $$
 +
 +Thus, the probability that the machine breaks down given that it underwent preventive maintenance is \( p(A|B) = 0.05 \), or 5%.
 +
 +
 +
tanszek/oktatas/techcomm/conditional_probability.1727693358.txt.gz · Last modified: 2024/09/30 10:49 by knehez