91Ó°ÊÓ

In probability, a complementary event is simply the opposite of an event happening. If you have an event \(A\), its complement \(A^c\) occurs if \(A\) does not occur. Consider this: if you have a bag with two red balls and three blue balls, the event of drawing a red ball is complementary to drawing a blue ball. They can't happen at the same time, making them complementary.For any event \(A\):

• Probability of \(A\): How likely \(A\) is to occur.
• Probability of \(A^c\): Calculated as \(1 - P(A)\).

To illustrate, if \(P(A) = 0.95\), then \(P(A^c) = 1 - 0.95 = 0.05\). This tells us the chance that event \(A\) does not occur.

When dealing with the probability of the intersection of events, you're essentially looking for the probability that multiple events happen at the same time. For two events \(A\) and \(B\), their intersection, denoted \(A \cap B\), occurs if both \(A\) and \(B\) occur simultaneously. Consider rolling two dice:

• Getting a 4 on the first die is event \(A\).
• Getting a 5 on the second die is event \(B\).

The intersection \(A \cap B\) is the event where both dice show these numbers at the same time.For independent events, where one event does not impact the other, you can calculate \(P(A \cap B)\) using the multiplication rule: \[ P(A \cap B) = P(A) \times P(B) \]This formula applies to any number of independent events, such as \(A \cap B \cap C\). In our scenario, for events \(A\), \(B\), and \(C\), we calculated:\[ P(A \cap B \cap C) = 0.95 \times 0.73 \times 0.62 = 0.42997\].

The multiplication rule is essential when finding the probability of multiple independent events occurring together. For events to be independent, the outcome of one does not affect the outcome of another.If you have two independent events, \(A\) and \(B\), the probability of both happening is:\[ P(A \cap B) = P(A) \times P(B) \]For three independent events \(A\), \(B\), and \(C\):

• Multiply the probabilities: \(P(A) \times P(B) \times P(C)\).
• This gives you \(P(A \cap B \cap C)\), the probability that all three occur together.

A practical tip: always check for independence first! For our specific exercise, because \(A\), \(B\), and \(C\) are independent, their joint probability was easily calculated using this rule. This gave us a result of \(0.42997\) for \(P(A \cap B \cap C)\).