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The prior probabilities for events \(A_{1}\) and \(A_{2}\) are \(P\left(A_{1}\right)=.40\) and \(P\left(A_{2}\right)=.60 .\) It is also known that \(P\left(A_{1} \cap A_{2}\right)=0 .\) Suppose \(P\left(B | A_{1}\right)=.20\) and \(P\left(B | A_{2}\right)=.05\). a. Are \(A_{1}\) and \(A_{2}\) mutually exclusive? Explain. b. Compute \(P\left(A_{1} \cap B\right)\) and \(P\left(A_{2} \cap B\right)\). c. Compute \(P(B)\). d. Apply Bayes' theorem to compute \(P\left(A_{1} | B\right)\) and \(P\left(A_{2} | B\right)\).

Short Answer

Expert verified
a. Yes, they are mutually exclusive. b. 0.08 and 0.03. c. 0.11. d. 0.727 and 0.273.

Step by step solution

01

Determine Mutual Exclusivity

Events are mutually exclusive if they cannot occur simultaneously. Given \(P(A_1 \cap A_2) = 0\), this implies that both events cannot happen at the same time, so \(A_1\) and \(A_2\) are mutually exclusive.
02

Calculate Joint Probability for A1 and B

The joint probability \(P(A_1 \cap B)\) can be calculated using the formula: \[P(A_1 \cap B) = P(B | A_1) \cdot P(A_1) \]. Substituting the given values, we get: \[P(A_1 \cap B) = 0.20 \cdot 0.40 = 0.08\].
03

Calculate Joint Probability for A2 and B

Similarly, the joint probability \(P(A_2 \cap B)\) is calculated using: \[P(A_2 \cap B) = P(B | A_2) \cdot P(A_2) \]. Substituting the given values, we have: \[P(A_2 \cap B) = 0.05 \cdot 0.60 = 0.03\].
04

Compute Probability of B

The probability of event \(B\) can be determined by the sum of joint probabilities: \[P(B) = P(A_1 \cap B) + P(A_2 \cap B)\]. Substituting the results from the previous steps: \[P(B) = 0.08 + 0.03 = 0.11\].
05

Use Bayes' Theorem for A1 Given B

Bayes' theorem states \(P(A_1 | B) = \frac{P(A_1 \cap B)}{P(B)}\). Plugging in values from our previous results gives: \[P(A_1 | B) = \frac{0.08}{0.11} \approx 0.727\].
06

Use Bayes' Theorem for A2 Given B

Similarly, using Bayes' theorem for \(P(A_2 | B)\): \[P(A_2 | B) = \frac{P(A_2 \cap B)}{P(B)}\]. Substitute the known quantities: \[P(A_2 | B) = \frac{0.03}{0.11} \approx 0.273\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mutual Exclusivity
In probability, mutual exclusivity refers to the situation where two events cannot happen at the same time. This is an important concept because it simplifies the calculations involved in probability problems. When two events are mutually exclusive, the probability of both occurring is zero. That is, the intersection of the two events is empty.

Let's consider the example where events \(A_1\) and \(A_2\) have prior probabilities \(P(A_1) = 0.40\) and \(P(A_2) = 0.60\). It is given that \(P(A_1 \cap A_2) = 0\). This indicates these two events are mutually exclusive because their intersection is a null event.

This means that:\[P(A_1 \text{ and } A_2) = 0\]
In simpler terms, \(A_1\) and \(A_2\) cannot happen at the same time, making them mutually exclusive events. Mutual exclusivity directly impacts how we compute probabilities in more complex scenarios.
Probability
Probability is the measure of the likelihood that an event will occur. It can range from 0, meaning the event is impossible, to 1, indicating it is certain. In the context of our example, the idea is to find out the likelihood of certain events occurring based on the given data.

For instance, given two events \(A_1\) and \(A_2\) with probabilities \(P(A_1) = 0.40\) and \(P(A_2) = 0.60\), we know these individual probabilities give us a starting point for further calculations. Additionally, with the help of conditional probabilities like \(P(B|A_1) = 0.20\) and \(P(B|A_2) = 0.05\), we can investigate how likely event \(B\) is when either \(A_1\) or \(A_2\) occurs.

Probability plays a crucial role in decision-making processes and statistical analysis. By breaking down each part of a probability question—such as determining exclusivity, conditional guarantees, or overall likelihood—we can better understand real-world situations and random processes.
Joint Probability
Joint probability refers to the probability of two events happening at the same time. It's an important aspect of probability theory, particularly when working with non-mutually exclusive events. In our example, despite events \(A_1\) and \(A_2\) being mutually exclusive, we are still interested in the joint probability when looking at event \(B\).

To calculate joint probabilities, we use formulas like \(P(A_1 \cap B) = P(B | A_1) \cdot P(A_1)\). Substituting the given values, we find:\[P(A_1 \cap B) = 0.20 \cdot 0.40 = 0.08\]
Likewise, for \(A_2\) and \(B\): \[P(A_2 \cap B) = 0.05 \cdot 0.60 = 0.03\]

These calculations show how multiple random events overlap, providing a more comprehensive view of an unpredictable scenario. Understanding joint probabilities aids in determining the combined likelihood of multiple outcomes, something essential for tasks such as risk assessment and strategic planning.

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Most popular questions from this chapter

In the city of Milford, applications for zoning changes go through a two-step process: a review by the planning commission and a final decision by the city council. At step 1 the planning commission reviews the zoning change request and makes a positive or negative recommendation concerning the change. At step 2 the city council reviews the planning commission's recommendation and then votes to approve or to disapprove the zoning change. Suppose the developer of an apartment complex submits an application for a zoning change. Consider the application process as an experiment. a. How many sample points are there for this experiment? List the sample points. b. Construct a tree diagram for the experiment.

A decision maker subjectively assigned the following probabilities to the four outcomes of an experiment: \(P\left(E_{1}\right)=.10, P\left(E_{2}\right)=.15, P\left(E_{3}\right)=.40,\) and \(P\left(E_{4}\right)=.20 .\) Are these probability assignments valid? Explain.

Jerry Stackhouse of the National Basketball Association's Dallas Mavericks is the best freethrow shooter on the team, making \(89 \%\) of his shots (ESPN website, July, 2008 ). Assume that late in a basketball game, Jerry Stackhouse is fouled and is awarded two shots. a. What is the probability that he will make both shots? b. What is the probability that he will make at least one shot? c. What is the probability that he will miss both shots? d. Late in a basketball game, a team often intentionally fouls an opposing player in order to stop the game clock. The usual strategy is to intentionally foul the other team's worst free-throw shooter. Assume that the Dallas Mavericks' center makes \(58 \%\) of his free-throw shots. Calculate the probabilities for the center as shown in parts (a), (b), and (c), and show that intentionally fouling the Dallas Mavericks' center is a better strategy than intentionally fouling Jerry Stackhouse.

The U.S. Census Bureau provides data on the number of young adults, ages \(18-24,\) who are living in their parents' home. \(^{1}\) Let \(M=\) the event a male young adult is living in his parents' home \(F=\) the event a female young adult is living in her parents' home If we randomly select a male young adult and a female young adult, the Census Bureau data enable us to conclude \(P(M)=.56\) and \(P(F)=.42\) (The World Almanac, 2006 ). The probability that both are living in their parents' home is .24 a. What is the probability at least one of the two young adults selected is living in his or her parents' home? b. What is the probability both young adults selected are living on their own (neither is living in their parents' home)?

An experiment with three outcomes has been repeated 50 times, and it was learned that \(E_{1}\) occurred 20 times, \(E_{2}\) occurred 13 times, and \(E_{3}\) occurred 17 times. Assign probabilities to the outcomes. What method did you use?

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