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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 \(P\left(B \text { ? } A_{1}\right)=.20\) 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, A1 and A2 are mutually exclusive. b. P(A1 ∩ B) = 0.08, P(A2 ∩ B) = 0.03. c. P(B) = 0.11. d. P(A1 | B) ≈ 0.727, P(A2 | B) ≈ 0.273.

Step by step solution

01

Determine if A1 and A2 are Mutually Exclusive

Events are mutually exclusive if they cannot occur at the same time, meaning their intersection is zero. Given, \(P(A_{1} \cap A_{2}) = 0\), it confirms that \(A_{1}\) and \(A_{2}\) are mutually exclusive events.
02

Compute P(A1 ∩ B)

Using the formula for conditional probability, \(P(A_{1} \cap B) = P(B | A_{1}) \times P(A_{1})\). Substitute the given values: \(P(A_{1} \cap B) = 0.20 \times 0.40 = 0.08\).
03

Compute P(A2 ∩ B)

Similarly, \(P(A_{2} \cap B) = P(B | A_{2}) \times P(A_{2})\). Substitute the given values: \(P(A_{2} \cap B) = 0.05 \times 0.60 = 0.03\).
04

Compute P(B)

Since \(P(B)\) can be calculated as the sum of the probabilities of \(B\) given either \(A_{1}\) or \(A_{2}\), use the formula: \(P(B) = P(A_{1} \cap B) + P(A_{2} \cap B)\). Therefore, \(P(B) = 0.08 + 0.03 = 0.11\).
05

Apply Bayes' Theorem to Compute P(A1 | B)

Bayes' Theorem states \(P(A_{1} | B) = \frac{P(A_{1} \cap B)}{P(B)}\). Substitute the known values: \(P(A_{1} | B) = \frac{0.08}{0.11} \approx 0.727\).
06

Apply Bayes' Theorem to Compute P(A2 | B)

Similarly, \(P(A_{2} | B) = \frac{P(A_{2} \cap B)}{P(B)}\). Substitute the known values: \(P(A_{2} | B) = \frac{0.03}{0.11} \approx 0.273\).

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Key Concepts

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

Mutually Exclusive Events
Mutually exclusive events are events that cannot happen at the same time. If one event occurs, the other cannot. An important feature of mutually exclusive events is that the intersection of these events is zero, meaning there is no overlap in their outcomes.
In mathematical terms, for two events, say event A and event B, to be mutually exclusive, it must hold that:
  • \( P(A \cap B) = 0 \)
In the exercise, we have events \( A_1 \) and \( A_2 \) such that \( P(A_1 \cap A_2) = 0 \). This confirms that the events \( A_1 \) and \( A_2 \) are indeed mutually exclusive.
Understanding whether events are mutually exclusive is crucial as it determines how probabilities are calculated when combining events.
For mutually exclusive events, the probability of either event occurring is simply the sum of their individual probabilities:
  • \( P(A \cup B) = P(A) + P(B) \)
Mutually exclusive events simplify the process of calculating probabilities because there is no overlap or shared probability between the events.
Conditional Probability
Conditional probability is the probability of an event occurring given that another event has already occurred. It provides a way to understand the likeliness of an event based on the occurrence of a previous event.
Mathematically, the conditional probability of event B given event A is written as \( P(B | A) \) and is calculated as:
  • \( P(B | A) = \frac{P(A \cap B)}{P(A)} \)
This formula states that to find the probability of B happening given A, we divide the probability of both A and B occurring by the probability of A occurring alone.
In the example exercise, conditional probabilities were utilized to derive \( P(A_1 \cap B) \) and \( P(A_2 \cap B) \), which translates to finding the interaction of each event A with event B:
  • \( P(A_1 \cap B) = P(B | A_1) \times P(A_1) = 0.20 \times 0.40 = 0.08 \)
  • \( P(A_2 \cap B) = P(B | A_2) \times P(A_2) = 0.05 \times 0.60 = 0.03 \)
Recognizing how conditional probability ties an event's outcome to another provides insight into dependent events, where outcomes are influenced by prior occurrences.
Intersection of Events
The intersection of events refers to the occurrence of two or more events at the same time, often denoted as \( A \cap B \). This is a key concept because it determines how likely it is for multiple events to coincide.
When working with intersections, it's essential to understand the relationships between these events—whether they are related or independent, as this impacts how the intersection is calculated.
For dependent events, where the occurrence of one affects the other, the formula for the intersection is:
  • \( P(A \cap B) = P(B | A) \times P(A) \)
This formula was applied in the original exercise problem to find the intersections \( P(A_1 \cap B) \) and \( P(A_2 \cap B) \).
When events are mutually exclusive (like \( A_1 \) and \( A_2 \) in our example), their intersection is zero:
  • \( P(A_1 \cap A_2) = 0 \)
Grasping the idea of intersections aids in accurately determining joint probabilities and paints a clearer picture of how events interact within probability theory.

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Most popular questions from this chapter

An oil company purchased an option on land in Alaska. Preliminary geologic studies assigned the following prior probabilities. \\[ \begin{aligned} P(\text { high-quality oil }) &=.50 \\ P(\text { medium-quality oil }) &=.20 \\ P(\text { no oil }) &=.30 \end{aligned} \\] a. What is the probability of finding oil? b. After 200 feet of drilling on the first well, a soil test is taken. The probabilities of finding the particular type of soil identified by the test follow. \\[ \begin{aligned} P(\text { soil } | \text { high-quality oil }) &=.20 \\ P(\text { soil } | \text { medium-quality oil }) &=.80 \\ P(\text { soil } | \text { no oil }) &=.20 \end{aligned} \\] How should the firm interpret the soil test? What are the revised probabilities, and what is the new probability of finding oil?

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