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Chapter 4: Problem 8

Basic Computation: Rules of Probability Given \(P\left(A^{c}\right)=0.8, P(B)=0.3\) \(P(B | A)=0.2:\) (a) Compute \(P(A \text { and } B).\) (b) Compute \(P(A \text { or } B).\)

Short Answer

Expert verified
(a) \( P(A \text{ and } B) = 0.04 \), (b) \( P(A \text{ or } B) = 0.46 \).

Step by step solution

01

Find P(A)

To find the probability of event A, we need to find the complement of the probability given. We know that the probability of the complement of A, denoted as \( P(A^c) \), is 0.8. Since \( P(A) + P(A^c) = 1 \), we can calculate \( P(A) \) as follows:\[ P(A) = 1 - P(A^c) = 1 - 0.8 = 0.2 \].
02

Compute P(A and B)

We use the rule for conditional probability, where \( P(B | A) = \frac{P(A ext{ and } B)}{P(A)} \). We have \( P(B | A) = 0.2 \) and we just calculated \( P(A) = 0.2 \). Thus,\[ P(A ext{ and } B) = P(B | A) \times P(A) = 0.2 \times 0.2 = 0.04 \].
03

Compute P(A or B)

To find the probability of either event A or event B occurring, use the formula: \( P(A ext{ or } B) = P(A) + P(B) - P(A ext{ and } B) \). Now substitute the known values:\[ P(A ext{ or } B) = 0.2 + 0.3 - 0.04 = 0.46 \].

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Key Concepts

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

Conditional Probability
Conditional probability helps us understand how the probability of one event can change based on the occurrence of another event. Here, it shows the likelihood of event B occurring given that A has already happened. This is expressed as \( P(B | A) \). In general, the formula for conditional probability is:
  • \( P(B | A) = \frac{P(A \text{ and } B)}{P(A)} \)
The conditional probability \( P(B | A) \) is 0.2 in our exercise, meaning that once event A occurs, event B has a 20% chance of occurring as well. We used this concept to determine \( P(A \text{ and } B) \), the probability of both A and B happening together by rearranging the conditional probability formula:
  • \( P(A \text{ and } B) = P(B | A) \times P(A) = 0.2 \times 0.2 = 0.04 \)
Understanding conditional probability is key to finding how related events impact one another in terms of their probabilities.
Complementary Events
In probability theory, each event has a complementary event, which is everything that is not part of the original event. If you know the probability of an event's complement, you can easily find the probability of the event itself.
  • The relationship is simple: \( P(A) + P(A^c) = 1 \).
Given \( P(A^c) = 0.8 \) in the exercise, we calculated the probability of event A by subtracting the probability of its complement from 1:
  • \( P(A) = 1 - P(A^c) = 1 - 0.8 = 0.2 \)
Complementary events cover all possible outcomes of a situation. They are useful because they often make it easier to calculate rather than dealing with the main event directly, especially if you have the information on the complement probability readily available.
Addition Rule in Probability
The addition rule is a fundamental principle used to compute the probability of either one of two events occurring. It helps us find \( P(A \text{ or } B) \), which is the probability that either event A occurs, or event B occurs, or both occur together.
  • The formula is: \( P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B) \).
In the exercise, we had:
  • \( P(A) = 0.2 \)
  • \( P(B) = 0.3 \)
  • \( P(A \text{ and } B) = 0.04 \)
Using these values, the calculation was:
  • \( P(A \text{ or } B) = 0.2 + 0.3 - 0.04 = 0.46 \)
This rule is pivotal in understanding how probabilities of multiple events intertwine, ensuring you're not overcounting the probability where both events occur.

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Most popular questions from this chapter

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According to a recent Harris Poll of adults with pets, the probability that the pet owner cooks especially for the pet either frequently or occasionally is 0.24. (a) From this information, can we conclude that the probability a male owner cooks for the pet is the same as for a female owner? Explain. (b) According to the poll, the probability a male owner cooks for his pet is 0.27 whereas the probability a female owner does so is 0.22. Let's explore how such probabilities might occur. Suppose the pool of pet owners surveyed consisted of 200 pet owners, 100 of whom are male and 100 of whom are female. Of the pet owners, a total of 49 cook for their pets. Of the 49 who cook for their pets, 27 are male and 22 are female. Use relative frequencies to determine the probability a pet owner cooks for a pet, the probability a male owner cooks for his pet, and the probability a female owner cooks for her pet.

Therapy: Alcohol Recovery The Eastmore Program is a special program to help alcoholics. In the Eastmore Program, an alcoholic lives at home but undergoes a two-phase treatment plan. Phase I is an intensive group-therapy program lasting 10 weeks. Phase II is a long-term counseling program lasting 1 year. Eastmore Programs are located in most major cities, and past data gave the following information based on percentages of success and failure collected over a long period of time: The probability that a client will have a relapse in phase I is \(0.27 ;\) the probability that a client will have a relapse in phase II is \(0.23 .\) However, if a client did not have a relapse in phase I, then the probability that this client will not have a relapse in phase II is 0.95. If a client did have a relapse in phase I, then the probability that this client will have a relapse in phase II is 0.70. Let \(A\) be the event that a client has a relapse in phase I and \(B\) be the event that a client has a relapse in phase II. Let \(C\) be the event that a client has no relapse in phase I and \(D\) be the event that a client has no relapse in phase II. Compute the following: (a) \(P(A), P(B), P(C),\) and \(P(D)\) (b) \(P(B | A)\) and \(P(D | C)\) (c) \(P(A \text { and } B)\) and \(P(C \text { and } D\) ) (d) \(P(A \text { or } B)\) (e) What is the probability that a client will go through both phase I and phase II without a relapse? (f) What is the probability that a client will have a relapse in both phase I and phase II? (g) What is the probability that a client will have a relapse in either phase I or phase I??

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