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Chapter 27: Problem 45

Let \(A\) and \(B\) be two events such that \(P(\overline{A \cup B})=\frac{1}{6}\), \(P(A \cap B)=\frac{1}{4}\) and \(P(\bar{A})=\frac{1}{4}\), where \(\bar{A}\) stands for complement of event \(A\). Then events \(A\) and \(B\) are [2005] (a) equally likely and mutually exclusive (b) equally likely but not independent (c) independent but not equally likely (d) mutually exclusive and independent

Short Answer

Expert verified
Events \(A\) and \(B\) are independent but not equally likely.

Step by step solution

01

Identify Relevant Formulas

To solve this problem, we need three probability concepts: complement rules, probability addition rules, and independence. We know if events are independent, then \( P(A \cap B) = P(A) \cdot P(B) \). Also, we have \( P(\bar{A}) = 1 - P(A) \) and \( P(\overline{A \cup B}) = 1 - P(A \cup B) \).
02

Calculate \(P(A)\)

Given \(P(\bar{A}) = \frac{1}{4}\), use \(P(A) = 1 - P(\bar{A})\) to find \(P(A)\). So, \(P(A) = 1 - \frac{1}{4} = \frac{3}{4}\).
03

Calculate \(P(A \cup B)\) Using Complement

We know \(P(\overline{A \cup B}) = \frac{1}{6}\), therefore, \(P(A \cup B) = 1 - P(\overline{A \cup B}) = 1 - \frac{1}{6} = \frac{5}{6}\).
04

Use Probability Addition Formula

The formula for the union of two events is \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). Plug in the known values: \(\frac{5}{6} = \frac{3}{4} + P(B) - \frac{1}{4}\).
05

Solve for \(P(B)\)

Rearrange the above equation to solve for \(P(B)\): \(P(B) = \frac{5}{6} - \frac{2}{4}\). Simplify the right side to get \(P(B) = \frac{1}{3}\).
06

Check Independence

For two events to be independent, \(P(A \cap B) = P(A) \cdot P(B)\). Check if \(\frac{1}{4} = \frac{3}{4} \cdot \frac{1}{3}\). The right side simplifies to \(\frac{3}{12} = \frac{1}{4}\), so they are independent.
07

Check if Events are Equally Likely

Two events are equally likely if \(P(A) = P(B)\). Since \(P(A) = \frac{3}{4}\) and \(P(B) = \frac{1}{3}\), they are not equally likely.

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Key Concepts

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

Probability principles
Probability is a measure of the likelihood that a particular event will occur. It relies on fundamental principles that help understand random events and their associated uncertainties. There are a few key principles to keep in mind:
  • Complement Rule: For any event, the probability that it does not occur is called the complement of the event. If the probability that event \(A\) occurs is \(P(A)\), the probability that \(A\) does not occur, denoted as \(\bar{A}\), is \(1 - P(A)\).
  • Addition Rule: This is used to determine the probability of the union of two events (\(A \cup B\)). For any two events \(A\) and \(B\), the probability that at least one of them occurs is \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
  • Multiplication Rule: This principle helps determine the probability that two independent events occur together. If events \(A\) and \(B\) are independent, their joint probability is \(P(A \cap B) = P(A) \cdot P(B)\).
Each of these principles provides a tool for calculating different aspects of probability, helping us solve various problems, like determining event likelihoods and dependencies.
Event independence
In probability theory, two events are considered independent if the occurrence of one event does not affect the probability of the other. This concept is critical when analyzing event relationships.
  • If events \(A\) and \(B\) are independent, it means \(P(A \cap B) = P(A) \cdot P(B)\).
  • Independence implies that events do not influence each other.
  • Checking independence often involves verifying this mathematical condition using given probabilities.
In practical terms, if you have determined that \(P(A \cap B) = \frac{1}{4}\) and then calculate \(P(A) \cdot P(B) = \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}\), you can confidently say that \(A\) and \(B\) are independent. Understanding this independence allows for easier calculations and predictions, as the relationship between events is clearly defined.
Probability formulas
There are key formulas in probability that help solve problems involving events and their likelihoods. These formulas allow us to compute probabilities systematically and with precision.
  • Complement Formula: \(P(\bar{A}) = 1 - P(A)\) helps us find the chance that an event does not happen.
  • Addition Formula: The formula \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) calculates the probability that either of two events (or both) happen.
  • Multiplication for Independence: \(P(A \cap B) = P(A) \cdot P(B)\) when \(A\) and \(B\) are independent shows joint probability.
These formulas are the building blocks of probability calculations. They are vital for determining specific outcomes and understanding how various events interact or coexist. When using these formulas, careful substitution of known values leads to a clear understanding of probability scenarios, as seen in solving the given exercise. Employing these tools effectively can simplify even complex probability problems.

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Most popular questions from this chapter

For an initial screening of an admission test, a candidate is given fifty problems to solve. If the probability that the candidate can solve any problem is \(\frac{4}{5}\), then the probability that he is unable to solve less than two problems is: [April 12, 2019 (II)] (a) \(\frac{201}{5}\left(\frac{1}{5}\right)^{49}\) (b) \(\frac{316}{25}\left(\frac{4}{5}\right)^{48}\) (c) \(\frac{54}{5}\left(\frac{4}{5}\right)^{49}\) (d) \(\frac{164}{25}\left(\frac{1}{5}\right)^{48}\)

A box contains 15 green and 10 yellow balls. If 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn is: [2017] (a) \(\frac{6}{25}\) (b) \(\frac{12}{5}\) (c) 6 (d) 4

Four persons can hit a target correctly with probabilities \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}\) and \(\frac{1}{8}\) respectively. If all hit at the target independently, then the probability that the target would be hit, is: \(\quad\) [April 09, 2019 (I)] (a) \(\frac{25}{192}\) (b) \(\frac{7}{32}\) (c) \(\frac{1}{192}\) (d) \(\frac{25}{32}\)

A dice is tossed 5 times. Getting an odd number is considered a success. Then the variance of distribution of success is (a) \(8 / 3\) (b) \(3 / 8\) (c) \(4 / 5\) (d) \(5 / 4\)

In a workshop, there are five machines and the probability of any one of them to be out of service on a day is \(\frac{1}{4}\) If the probability that at most two machines will be out of service on the same day is \(\left(\frac{3}{4}\right)^{3} k\), then \(k\) is equal to: [Jan. 7, 2020 (II)] (a) \(\frac{17}{8}\) (b) \(\frac{17}{4}\) (c) \(\frac{17}{2}\) (d) 4
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