91Ó°ÊÓ

In probability theory, events are considered independent when the occurrence of one event does not affect the occurrence of another. In the context of our exercise, the failures of systems A and B are independent events. This means that whether system A fails or not has no bearing on the failure of system B and vice versa.
To check if events are independent, you can verify that the joint probability of the events occurring (or not occurring) is equal to the product of their individual probabilities. For example, if event X and event Y are independent, then:
\[ P(X \text{ and } Y) = P(X) \times P(Y) \]
In our problem, if the probability of system A failing is 0.12, and the probability of system B failing is 0.07, the probability that both systems fail is:
\[P(A^c \text{ and } B^c) = P(A^c) \times P(B^c) = 0.12 \times 0.07 = 0.0084 \]

Joint probability refers to the probability of two (or more) events occurring simultaneously. In simpler terms, it is the likelihood of multiple conditions being satisfied at the same time.
In our scenario, we are interested in the joint probability of both Systems A and B failing which is the network failing condition.
The formula to find this, given that the events are independent, is to multiply the individual probabilities of the events. So for events A and B, it is calculated as follows:
\[ P(A^c \text{ and } B^c) = P(A^c) \times P(B^c) \]
For our example:
* Probability that system A fails: 0.12
* Probability that system B fails: 0.07

When calculated the joint probability is:
\[ P(A^c \text{ and } B^c) = 0.12 \times 0.07 = 0.0084 \]
This means that there is a 0.84% chance that both systems fail simultaneously, leading to a network failure.

Understanding complementary probability is essential for calculating the likelihood of an event not happening based on the probability of it happening.The complementary probability of an event A, denoted as \( P(A^c) \), is the probability that event A does not occur. This is calculated as:
\[ P(A^c) = 1 - P(A) \]
Applying this to our problem, we have:
* Probability that system A functions (\( P(A) \)): 0.88
* Hence Probability that system A fails (\( P(A^c) \)): \( 1 - 0.88 = 0.12 \)
Similarly for system B:
* Probability that system B functions (\( P(B) \)): 0.93
* Hence Probability that system B fails (\( P(B^c) \)): \( 1 - 0.93 = 0.07 \)
Understanding complementary probability helps in calculating the likelihood of events indirectly by focusing on their non-occurrence. In this case, calculating \( P(A^c) \) and \( P(B^c) \) was crucial to finding the joint probability of a network failure.