Problem 56 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(A\) and \(B\) are events of an experiment, then $$P(A \cap B)=P(A \mid B) \cdot P(B)=P(B \mid A) \cdot P(A)$$

Step-by-Step Solution

Answer

The statement is True. It can be verified using the definition of conditional probability: $$P(A \cap B) = P(A \mid B) \cdot P(B) = P(B \mid A) \cdot P(A)$$

Step 1: Definition of Conditional Probability

According to the definition of conditional probability, the probability of event \(A\) happening given event \(B\) already happened is given by: $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ Similarly, the probability of event \(B\) happening given event \(A\) already happened is given by: $$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$$

Step 2: Determine if the statement is true or false

We have the given statement: $$P(A \cap B)=P(A \mid B) \cdot P(B)=P(B \mid A) \cdot P(A)$$ We will first test if the following expression holds true: $$P(A \cap B)=P(A \mid B) \cdot P(B)$$ Using the definition of conditional probability, we can rewrite the expression as: $$P(A \cap B) = \frac{P(A \cap B)}{P(B)} \cdot P(B)$$ Since the \(P(A \cap B)\) and \(P(B)\) terms on the right-hand side cancel each other, we obtain: $$P(A \cap B) = P(A \cap B)$$ Now, we will test if the second part of the statement also holds true: $$P(A \cap B)=P(B \mid A) \cdot P(A)$$ Using the definition of conditional probability, we can rewrite the expression as: $$P(A \cap B) = \frac{P(A \cap B)}{P(A)} \cdot P(A)$$ Since the \(P(A \cap B)\) and \(P(A)\) terms on the right-hand side cancel each other, we obtain: $$P(A \cap B) = P(A \cap B)$$

Step 3: Conclusion

Since both of the expressions in the statement hold true, we can conclude that the given statement is True. $$P(A \cap B)=P(A \mid B) \cdot P(B)=P(B \mid A) \cdot P(A)$$

Key Concepts

Probability of Events

Understanding probability of events is crucial for analysing outcomes in experiments, whether flipping a coin or predicting weather. Probability measures the likelihood of an event occurring and is expressed as a number between 0 and 1, where 0 indicates an impossible event, and 1 indicates a certainty.

For any event 'A', the probability, denoted by P(A), is calculated as the number of favorable outcomes divided by the total number of possible outcomes. It's important for students to recognize that each outcome must be equally likely for this basic formula to hold true.

For instance, the probability of rolling a die and getting a 3 can be calculated as: \[ P(\text{{roll a 3}}) = \frac{1}{6} \]
This represents one favorable outcome (rolling a 3) over the six total possible outcomes.

Intersection of Events

The concept of the intersection of events is pivotal when examining the occurrence of two events at the same time. For events A and B, their intersection, denoted by \(A \cap B\), represents the set of all outcomes that are common to both A and B. Put simply, it asks: What are the chances both A and B happen together?

To put it into context, consider drawing a card from a standard deck and wanting both a King (event A) and a Heart (event B). The intersection of these events would be drawing the King of Hearts, since that is the outcome common to both being a King and being a Heart.

Calculating Intersection Probability

The probability of the intersection of two events is given by the product of the probability of one event and the conditional probability of the second event given that the first one has occurred:\[ P(A \cap B) = P(A) \times P(B \mid A) \]
If A and B are independent events, meaning the occurrence of one does not affect the probability of the other, this simplifies to:\[ P(A \cap B) = P(A) \times P(B) \]

Probability Theory

Probability theory is a branch of mathematics that deals with the analysis of random events. The cornerstone of probability theory is the rules that govern probabilities, especially the rule for combining the probabilities of events in a precise way, ensuring logical consistency and predictability in an otherwise uncertain world.

The statement we assessed in the exercise, \[P(A \cap B) = P(A \mid B) \cdot P(B)\], encapsulates the fundamental relationship between joint and conditional probabilities. This relationship is rooted in probability theory and enables us to dissect complex events into more manageable parts for analysis.

One of the crucial aspects of probability theory that helps in solving many real-world problems is the concept of independence. If the occurrence of one event doesn't influence the occurrence of another, the two events are considered independent, a fact used when calculating probabilities in many practical situations such as risk assessment and decision-making processes.