Conditional probability
Non-independent events
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.
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).
Notation: \(p(A | B) \)
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:
$$ p(A \cap B) = p(A|B) p(B) $$
1.) \( p(A \cap B) \): This represents the probability that both events A and B occur simultaneously, which is also known as the probability of their intersection.
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 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%.