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--- tanszek:oktatas:techcomm:conditional_probability [2024/08/26 17:54] – knehez
+++ tanszek:oktatas:techcomm:conditional_probability [2024/10/13 17:27] (current) – kissa
@@ Line 1: / Line 1: @@
-==== Conditional probability ====
+===== Conditional probability =====
-How can we calculate the result in a case where two events are not independent. It means that, if one event occurs it will directly affect the probability for the other event?
+=== Non-independent events ===
-If event A and B are those kind of complex events which will not exclude each other. In this case we have a so-called conditional probability (event A affects event B).
+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.
-Notation: \(p(A | B) \)
+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).
-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).
+**Notation**: \(p(A | B) \)
-$$ p(A|B) = \frac{k_{AB}}{k_b} = \frac{\frac{k_{AB}}{k}}{\frac{k_{B}}{k_b}} = \frac{p(A \cap B)}{p(B)} $$
+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).
+$$ p(A|B) = \frac{k_{AB}}{k_b} = \frac{\frac{k_{AB}}{k}}{\frac{k_{B}}{k}} = \frac{p(A \cap B)}{p(B)} $$
 So we can get to the conclusion:
 $$ p(A \cap B) = p(A|B) p(B) $$
+.) \( p(A \cap B) \):
+This represents the probability that both events
+A and B occur simultaneously, which is also known as the probability of their intersection.
+.) \(  p(A|B) \): 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.
+**What the Formula Says?**
+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 1**:
+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:
+^ Measurement Result ^ Quantity ^
+| Faultless \((H)\) | 162 |
+| The length 𝐿 is faulty \((A)\) | 10 |
+| The diameter 𝐷 is faulty \((B)\) | 12 |
+| Both dimensions are faulty \( ( A \cap B ) \) | 4 |
+| Only the length 𝐿 is faulty \((C)\) | 6 |
+| Only the diameter 𝐷 is faulty \((D)\) | 8 |
+**Question 1**: What are the probabilities of events \(A\) and \(B\)?
+The probability of the event "length" is faulty" \((A)\) is:
+$$ p(A) = \frac{10}{180} = 0.05555 $$
+The probability of the event "diameter" is faulty" \((B)\) is:
+$$ p(B) = \frac{12}{180} = 0.06666 $$
+**Question 2**: What is the probability that both dimensions are faulty?
+$$ p(A \cap B) = \frac{4}{180} = 0.0222 $$
+**Question 3**: What is the probability that a shaft's length is faulty, given that its diameter is already faulty?
+The conditional probability of both events occurring can be calculated using the definition:
+$$ p(A \mid B) = \frac{\text{both dimensions are faulty}}{\text{diameter is faulty}} = \frac{4}{12} = 0.3333 $$
+Since this does not match with the product \( p(A) \cdot p(B)\), we can conclude that the two events are **not independent**!
+Thus, the joint probability can also be calculated differently:
+$$ p(A \cap B) = p(A \mid B) \cdot p(B) = 0.333 \cdot 0.0666 = 0.02222 $$
+The probability of event \(C\) is:
+$$ p(C) = \frac{6}{180} = 0.0333 $$
+The probability of event \(D\) is:
+$$ p(D) = \frac{8}{180} = 0.0444 $$
+**Question 4**: What is the probability of defective production?
+$$ p(H) = \frac{180 - 162}{180} = \frac{18}{180} = 0.1 $$
+Alternatively, we can calculate it as:
+$$ p(A \cup B) = p(A) + p(B) - p(A \cap B) = 0.0555 + 0.0666 - 0.0222 = 0.1 $$
+or
+$$ p(A \cup B \cup E) = 0.0333 + 0.0444 + 0.0222 = 0.1 $$
+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%.