Question

Suppose that \(P(A)=.4\) and \(P(A \cap B)=.12\) a. Find \(P(B \mid A)\). b. Are events \(A\) and \(B\) mutually exclusive? c. If \(P(B)=.3,\) are events \(A\) and \(B\) independent?

Short answer

Question: Determine the probability of event B given event A, and state if events A and B are mutually exclusive and independent. Solution: a. The probability of B given A is 0.3. b. Events A and B are not mutually exclusive. c. Events A and B are independent.

Step-by-step solution

Recall the formula for conditional probability

The formula for conditional probability is given by: \(P(B \mid A) = \frac{P(A \cap B)}{P(A)}\)

Find the probability of B given A

We are given \(P(A) = 0.4\) and \(P(A \cap B) = 0.12\). Let's use the formula from step 1 to find \(P(B \mid A)\). \(P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.12}{0.4} = 0.3\)

Determine if events A and B are mutually exclusive

Two events are mutually exclusive if their intersection is an empty set, i.e., they cannot both happen at the same time. Mathematically: \(A\) and \(B\) are mutually exclusive if \(P(A \cap B) = 0\) Since \(P(A \cap B) = 0.12\), events A and B are not mutually exclusive.

Recall the formula for independence

Two events are independent if the probability of their intersection is equal to the product of their probabilities, i.e., \(P(A \cap B) = P(A)P(B)\).

Determine if events A and B are independent

We are given \(P(A) = 0.4\), \(P(A \cap B) = 0.12\), and \(P(B) = 0.3\). Let's apply the formula for independence: \(P(A)P(B) = 0.4 \times 0.3 = 0.12\) Since \(P(A \cap B) = P(A)P(B)\), events A and B are independent. To summarize: a. The probability of B given A is \(0.3\). b. Events A and B are not mutually exclusive. c. Events A and B are independent.

Key concepts

Mutually Exclusive Events

Understanding mutually exclusive events is crucial in the study of probability. These are events that cannot occur at the same time. In other words, if one event happens, the other cannot. An easy example to grasp this concept with is flipping a coin. The outcome can either be heads or tails, but not both simultaneously. If you're looking at two events, say event A and event B, they are mutually exclusive if the intersection of these events, denoted as \(P(A \cap B)\), is zero. In the original exercise, we found that \(P(A \cap B)\) is not zero but 0.12, thus, the events A and B are not mutually exclusive events because there is a scenario where both can happen together.

In real-life terms, imagine you're at a buffet and you can choose either a slice of cake or a scoop of ice cream. If you're not allowed to have both, choosing the cake means you cannot have the ice cream, and vice versa, making these two options mutually exclusive.

Independent Events

Moving on to independent events, these are quite different from mutually exclusive events. Independence in probability suggests that the occurrence of one event does not impact the probability of the occurrence of another event. Think about rolling a die and flipping a coin. Whether the die shows a six has no effect on whether the coin lands tails up. Technically, we say that two events, A and B, are independent if the probability of their intersection \(P(A \cap B)\) is equal to the product of their individual probabilities: \(P(A) \times P(B)\). Using the numbers from our exercise, we calculated that \(P(A)P(B)\) equals 0.12, which is the same as \(P(A \cap B)\), confirming that events A and B are independent events.

This is like saying buying a lottery ticket each day has no influence on winning (assuming each draw is independent and the outcome of one doesn't affect the others), so your chance of winning on any given day is always the same regardless of past purchases.

Probability Intersection

Lastly, let's discuss the concept of probability intersection, which was fundamental in our exercise. The intersection of events A and B, represented as \(A \cap B\), describes the outcomes that are common to both events. It is the 'overlap' of the two events. When you calculate the probability of this intersection, \(P(A \cap B)\), you're essentially computing the likelihood that both events occur simultaneously. In the context of our exercise, we found out that \(P(A \cap B)\) was 0.12. Given the provided probabilities for A and B, this number was key to help us deduce both the conditional probability of B given A and the independence of the two events.

An everyday example might involve survey data, where you are looking at the number of people who drink coffee and also enjoy classical music. The intersection here would represent people who do both - drink coffee and enjoy classical music.