Question

Suppose that \(P(A)=.3\) and \(P(B)=.4\) a. If \(P(A \cap B)=.12\) are \(A\) and \(B\) independent? Justify your answer. b. If \(P(A \cup B)=.7\) what is \(P(A \cap B)\) ? Justify your answer. c. If \(A\) and \(B\) are independent, what is \(P(A \mid B)\) ? d. If \(A\) and \(B\) are mutually exclusive, what is \(P(A \mid B) ?\)

Short answer

If so, find the conditional probability of A given B. If they are mutually exclusive, find the conditional probability of A given B. Answer: Yes, events A and B are independent. The conditional probability of A given B is 0.3. If they were mutually exclusive, the conditional probability of A given B would be 0.

Step-by-step solution

a. Checking the independence of events A and B

To check if events A and B are independent, we need to verify if the probability of their intersection (A ∩ B) is equal to the product of their individual probabilities. Let's use the given probabilities: $$P(A \cap B) = 0.12$$ $$P(A) \times P(B) = 0.3 \times 0.4 = 0.12$$ Since the probability of their intersection is equal to the product of their individual probabilities, events A and B are independent.

b. Finding the probability of the intersection between A and B, given their union

We have the probability of the union between A and B, which is 0.7, and we need to find the probability of their intersection, using the formula: $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ Plug in the given probabilities and solve for the intersection probability: $$0.7 = 0.3 + 0.4 - P(A \cap B)$$ $$P(A \cap B) = 0.3 + 0.4 - 0.7$$ $$P(A \cap B) = 0$$

c. Finding the conditional probability \(P(A|B)\), given the independence of A and B

If events A and B are independent, we can find the conditional probability \(P(A|B)\) using the formula: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ Since we have already found the probabilities of A and B, and their independence, we can plug in the values to get the conditional probability: $$P(A|B) = \frac{0.12}{0.4} = 0.3$$

d. Finding the conditional probability \(P(A|B)\), given the mutual exclusiveness of A and B

When A and B are mutually exclusive, it means that they cannot both happen at the same time, which means that the intersection probability is 0: $$P(A \cap B) = 0$$ So, the conditional probability will also be 0, because there is no chance for A to happen given B: $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0}{0.4} = 0$$

Key concepts

Independence of Events

In probability theory, two events, say Event A and Event B, are considered independent if the occurrence of one does not affect the occurrence of the other.

This means that knowing Event A has happened tells you nothing about the likelihood of Event B happening. The mathematical way to express this independence is through the equation:

• \( P(A \cap B) = P(A) \times P(B) \)

This formula says the probability of both events happening together (their intersection) should be equal to the product of their individual probabilities.

Our exercise gives us numbers to check this. With \(P(A) = 0.3\), \(P(B) = 0.4\), and \(P(A \cap B) = 0.12\), we multiply \(0.3\) and \(0.4\), which indeed gives \(0.12\).

This confirms A and B as independent events. Remember, independence in probabilities tells us how two events relate, but it's logical to think through real-world implications as they can help solidify these abstract ideas.

Conditional Probability

Conditional probability is all about finding the probability of an event occurring given that another event has already occurred. This concept allows us to calculate how one event can affect another.

The formula to calculate the conditional probability of event A given event B is:

• \(P(A|B) = \frac{P(A \cap B)}{P(B)}\)

In our step-by-step solution, we figure out \(P(A|B)\) when A and B are independent. Since independence in this scenario confirms \(P(A \cap B) = P(A) \times P(B)\), it simplifies our work.

Plugging in the numbers, \(P(A|B) = \frac{0.12}{0.4} = 0.3\), which matches \(P(A)\).

Conditional probability becomes quite different when events are mutually exclusive. Because A and B cannot happen together in such a case, the conditional probability \(P(A|B)\) turns out to be 0, as shown in the last step of the exercise.

Mutually Exclusive Events

Mutually exclusive events are ones that cannot both happen at the same time. When one occurs, the other cannot. A simple everyday example is flipping a coin; it can't land on both heads and tails simultaneously.

In mathematical terms, mutually exclusive events A and B are described as:

• \(P(A \cap B) = 0\)

This means the probability of both events happening at the same time is zero.

In our exercise, if A and B are mutually exclusive, we find that \(P(A \cap B) = 0\).

This affects the conditional probability because \(P(A|B)\), calculated as \(\frac{P(A \cap B)}{P(B)}\), also results in zero. This means if B happens, A cannot happen, and vice versa.

Though it seems contrasting, understanding mutually exclusive events helps in differentiating situations where events can't coincide and realizing their implications in various probability scenarios.