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--- tanszek:oktatas:techcomm:conditional_probability [2024/09/30 10:51] – knehez
+++ tanszek:oktatas:techcomm:conditional_probability [2024/10/13 17:27] (current) – kissa
@@ Line 11: / Line 11: @@
 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:
@@ Line 27: / Line 27: @@
 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 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:
@@ Line 38: / Line 38: @@
 | Both dimensions are faulty \( ( A \cap B ) \) | 4 |
 | 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 \(B\)?
@@ Line 60: / Line 60: @@
 $$ p(A \mid B) = \frac{\text{both dimensions are faulty}}{\text{diameter is faulty}} = \frac{4}{12} = 0.3333 $$
-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:
@@ Line 87: / Line 87: @@
 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%.