91Ó°ÊÓ

Conditional probability helps us understand the likelihood of an event occurring, given that another event has already happened. In probability theory, this is a crucial concept, especially when the outcome of one event influences the outcome of another.

The mathematical expression for conditional probability is \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \]where \(P(A | B)\) is the probability of event A occurring given that B has occurred. The intersection \(P(A \cap B)\) represents the probability of both A and B happening together.

In our example, to determine \(P(A_2 | A_3)\), or the probability of liking vehicle #2, given the individual also likes vehicle #3, we use the formula and plug in the provided values: \[P(A_2 | A_3) = \frac{P(A_2 \cap A_3)}{P(A_3)} = \frac{0.40}{0.70} = \frac{4}{7}\]This calculation shows that the probability is approximately 0.5714, indicating a 57.14% chance the person will like vehicle #2, given they like vehicle #3.

Two events are said to be independent if the occurrence of one does not affect the probability of the occurrence of the other. This means that knowing the outcome of one event provides no information about the other.

Mathematically, events A and B are independent if:\[ P(A \cap B) = P(A) \times P(B) \]or, equivalently,\[ P(A | B) = P(A) \]In the exercise, the events \(A_2\) and \(A_3\), liking vehicles #2 and #3, need to be checked for independence. The product \(P(A_2) \times P(A_3) = 0.65 \times 0.70 = 0.455\) does not equal \(P(A_2 \cap A_3) = 0.40\). Therefore, these events are not independent.

Additionally, \(P(A_2 | A_3)\), calculated as \(\frac{4}{7} \approx 0.5714\), is not equal to \(P(A_2) = 0.65\), confirming that events \(A_2\) and \(A_3\) are not independent.

The inclusion-exclusion principle is a useful tool in probability to find the union probability of multiple events. It corrects for the over-counting that occurs when adding up probabilities of individual events.

For two events, the principle is expressed as:\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]This formula helps to precisely calculate the probability that either or both events occur.

In our example, to find the probability of an individual liking either vehicle #2 or #3, \(P(A_2 \cup A_3)\), we use:\[ P(A_2 \cup A_3) = P(A_2) + P(A_3) - P(A_2 \cap A_3) = 0.65 + 0.70 - 0.40 = 0.95 \]This gives a 95% probability, showing how the principle accounts for overlaps by subtracting the intersection probability, ensuring accuracy in union calculations. The same principle aids in evaluating conditional probabilities, enhancing the understanding of event interactions.