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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?
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).
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).
$$ p(A|B) = \frac{k_{AB}}{k_b} = \frac{\frac{k_{AB}}{k}}{\frac{k_{B}}{k_b}} = \frac{p(A \cap B)}{p(B)} $$
So we can get to the conclusion:
$$ p(A \cap B) = p(A|B) p(B) $$
1.) \( p(A \cap B) \): 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.
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.
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.
Example:
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:
Measurement Result | Quantity |
---|---|
Faultless \((H)\) | 162 |
The length ๐ฟ is faulty \((A)\) | 10 |
The diameter ๐ท is faulty \((B)\) | 12 |
Both dimensions are faulty \( ( A \cap B ) \) | 4 |
Only the length ๐ฟ is faulty \((C)\) | 6 |
Only the diameter ๐ท is faulty \(D)\) | 8 |
Question: What are the probabilities of events \(A\) and \(B\):