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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 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 \( p(A) p(B)\), 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 \( E = A \cap B \)