Problem 121 - Applied Statistics and Probability for Engineers

Question

Suppose that \(P(A \mid B)=0.4\) and \(P(B)=0.5 .\) Determine the following: (a) \(P(A \cap B\) (b) \(P\left(A^{\prime} \cap B\right)\)

Step-by-Step Solution

Answer

(a) 0.2, (b) 0.3

Step 1: Understand the Given Information

We are given the conditional probability \(P(A \mid B) = 0.4\) and the probability \(P(B) = 0.5\). We need to find the joint probability \(P(A \cap B)\) in part (a) and \(P(A' \cap B)\) in part (b).

Step 2: Use the Conditional Probability Formula

The formula for conditional probability is given by \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\). We need to use this formula to solve for \(P(A \cap B)\).

Step 3: Solve for \(P(A \cap B)\)

Using the formula \(P(A \mid B) = \frac{P(A \cap B)}{P(B)}\) and substituting the given values, we get: \[0.4 = \frac{P(A \cap B)}{0.5}\] By solving for \(P(A \cap B)\), we find: \[P(A \cap B) = 0.4 \times 0.5 = 0.2\].

Step 4: Use Probability Properties for Complementary Events

To find \(P(A' \cap B)\), use the property that \(P(A' \cap B) = P(B) - P(A \cap B)\). This is because the probability of event \(B\) can be divided into \(P(A \cap B)\) and \(P(A' \cap B)\).

Step 5: Solve for \(P(A' \cap B)\)

Now substitute the known values: \[P(A' \cap B) = 0.5 - 0.2 = 0.3\].

Key Concepts

Joint Probability

Joint probability allows us to understand the probability of two events occurring at the same time. When calculating joint probability, we are often interested in the overlap between two events, say, events A and B, which is symbolized by \( P(A \cap B) \).
To calculate this, one useful tool is conditional probability. This is especially helpful when you are given \( P(A \mid B) \), which is the probability of A occurring given that B has occurred. You can find the joint probability using the formula:

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

Given that \( P(A \mid B) = 0.4 \) and \( P(B) = 0.5 \), the calculation goes as follows:
Just multiply these values to get:
\[ P(A \cap B) = 0.4 \times 0.5 = 0.2 \] This result gives us a clear understanding of the likelihood that both events A and B occur at the same time.

Complementary Events

Complementary events are fascinating because they give insight into the complete range of outcomes in a probability scenario. If event A is likely to occur, then its complement, denoted \( A' \), is the event that A does not occur.
When dealing with probabilities, the complement rule states that the sum of the probabilities of an event and its complement is 1:

• \( P(A) + P(A') = 1 \)

In our context, when you're asked about \( P(A' \cap B) \), or the probability of not A and B occurring, you can use the decomposition of probabilities over B. Break down \( P(B) \) as the sum:

• \( P(B) = P(A \cap B) + P(A' \cap B) \)

Given that \( P(B) = 0.5 \) and \( P(A \cap B) = 0.2 \), simply solve for \( P(A' \cap B) \) by subtracting the known joint probability:

• \( P(A' \cap B) = 0.5 - 0.2 = 0.3 \)

Thus, \( P(A' \cap B) = 0.3 \) reflects the probability of B occurring without A.

Probability Properties

Understanding probability properties is fundamental as these principles provide the framework to solve complex problems.
Here are a few basic properties that help simplify most probability questions:

• Addition Rule: If you're dealing with events that cover all possible outcomes – let's say event A and its complement \( A' \), their probabilities will sum up to 1. This rule holds as \( P(A) + P(A') = 1 \).
• Multiplication Rule: For independent events, the probability of both occurring is the product of their probabilities. But when dependency exists, such as in conditional probability, the formula changes to \( P(A \cap B) = P(A \mid B) \times P(B) \).
• Decomposition of Events: If you know the probability of an event and its joint component, their difference leads to the complementary condition. Like finding \( P(A' \cap B) \) from \( P(B) \) and \( P(A \cap B) \).

Utilizing these properties makes handling and interpreting probabilities more intuitive. Armed with these concepts, students can tackle a broad range of probability problems with confidence.