tanszek:oktatas:techcomm:conditional_probability
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| Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
| tanszek:oktatas:techcomm:conditional_probability [2024/09/30 12:30] โ knehez | tanszek:oktatas:techcomm:conditional_probability [2024/10/13 17:27] (current) โ kissa | ||
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| | Both dimensions are faulty \( ( A \cap B ) \) | 4 | | | Both dimensions are faulty \( ( A \cap B ) \) | 4 | | ||
| | Only the length ๐ฟ is faulty \((C)\) | 6 | | | Only the length ๐ฟ is faulty \((C)\) | 6 | | ||
| - | | Only the diameter ๐ท is faulty \(D)\) | 8 | | + | | Only the diameter ๐ท 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 \(B\)? | ||
| Line 60: | Line 60: | ||
| $$ p(A \mid B) = \frac{\text{both dimensions are faulty}}{\text{diameter is faulty}} = \frac{4}{12} = 0.3333 $$ | $$ p(A \mid B) = \frac{\text{both dimensions are faulty}}{\text{diameter is faulty}} = \frac{4}{12} = 0.3333 $$ | ||
| - | 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: | ||
tanszek/oktatas/techcomm/conditional_probability.1727699425.txt.gz ยท Last modified: 2024/09/30 12:30 by knehez