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Let \(\mathscr{S}\) be the sample space for an experiment \(\mathscr{E}\) and let \(A, B, C \subseteq \mathscr{P}\). If events \(A, B\) are independent, events \(A, C\) are disjoint, and events \(B, C\) are independent, find \(\operatorname{Pr}(B)\) if \(\operatorname{Pr}(A)=0.2, \operatorname{Pr}(C)=0.4\), and \(\operatorname{Pr}(A \cup B \cup C)=0.8\).

Short Answer

Expert verified
\(\operatorname{Pr}(B) = 0.5

Step by step solution

01

Identify the given data

The company is given that \(\operatorname{Pr}(A)=0.2, \operatorname{Pr}(C)=0.4, \operatorname{Pr}(A \cup B \cup C)=0.8\) and that events A, B and B, C are independent, and A, C are disjoint.
02

Understand the meaning of independence and disjointness

If A and B are independent, the probability of both occurring is the product of the probabilities of each occurring (\( \operatorname{Pr}(A \cap B) = \operatorname{Pr}(A) \operatorname{Pr}(B) \)). If A and C are disjoint, they cannot happen at the same time, thus the probability of both occurring is zero (\( \operatorname{Pr}(A \cap C) = 0\))
03

Express the probability of the union of all events

The probability of the union of A, B, and C can be expressed as \(\operatorname{Pr}(A \cup B \cup C) = \operatorname{Pr}(A) + \operatorname{Pr}(B) + \operatorname{Pr}(C) - \operatorname{Pr}(A \cap B) - \operatorname{Pr}(A \cap C) - \operatorname{Pr}(B \cap C) + \operatorname{Pr}(A \cap B \cap C)\). However, since A and C are disjoint, \(\operatorname{Pr}(A \cap C) = 0\), and since B and C are independent and A and B are independent, we know that \( \operatorname{Pr}(A \cap B) = \operatorname{Pr}(A) \operatorname{Pr}(B) \) and \( \operatorname{Pr}(B \cap C) = \operatorname{Pr}(B) \operatorname{Pr}(C)\). Also because A and C are disjoint, B must be also disjoint with A and C, therefore \( \operatorname{Pr}(A \cap B \cap C) = 0\)
04

Use the given information to determine Pr(B)

Substitute the given values and the findings from step 3 into the equation from step 3 to solve for \( \operatorname{Pr}(B)\), \(\operatorname{Pr}(B) = [ \operatorname{Pr}(A \cup B \cup C) - \operatorname{Pr}(A) - \operatorname{Pr}(C) ] / {1 - \operatorname{Pr}(A) - \operatorname{Pr}(C)} = (0.8 - 0.2 - 0.4) / (1 - 0.2 - 0.4)\)

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Key Concepts

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

Independent Events
In probability theory, the concept of independent events is crucial. Two events, say event A and event B, are considered independent if the occurrence of one event does not affect the occurrence of the other. Mathematically, this relationship is expressed as:\[ \operatorname{Pr}(A \cap B) = \operatorname{Pr}(A) \operatorname{Pr}(B) \]This formula implies that the probability of both events happening together is simply the product of their individual probabilities. This concept is key in many statistical calculations, as it simplifies the complexity involved in calculating probabilities involving multiple events. When dealing with independent events, always ensure that the given situations respect this principle, and use it to find other probabilities as needed.
Disjoint Events
Disjoint events, also known as mutually exclusive events, are events that cannot occur at the same time. This means if event A happens, event C cannot happen, and vice versa. In terms of probability, for two disjoint events A and C, the probability of both events occurring simultaneously is zero:\[ \operatorname{Pr}(A \cap C) = 0 \]Disjoint events are straightforward because they simplify calculations by eliminating the possibility of overlap. For example, when summing probabilities of disjoint events, the overlap term is zero, making probabilities additive:\[ \operatorname{Pr}(A \text{ or } C) = \operatorname{Pr}(A) + \operatorname{Pr}(C) \]Remember, if you have disjoint events, always check if there are such zero-overlap areas in probability problems.
Sample Space
The sample space in probability is the set of all possible outcomes of an experiment. It provides the universal set from which events (subsets of outcomes) can be defined. For example, if you are rolling a six-sided die, the sample space, denoted by the symbol \(\mathscr{S}\), would be {1, 2, 3, 4, 5, 6}. Understanding the sample space is foundational because it allows us to frame events and calculate their probabilities. Every probability problem starts here, ensuring you account for all possible outcomes:
  • Defines the boundaries of what can happen.
  • Informs the calculation of event probabilities as fractions of the sample space.
Always begin by identifying your sample space as you gear up to solve probability questions.
Union of Events
The union of events in probability refers to the occurrence of at least one of the events happening. For example, the union of events A, B, and C, denoted as \(A \cup B \cup C\), means that at least one of these events occurs. The probability of the union addresses this occurrence. To calculate this, you often use the principle of inclusion-exclusion to correct for over-counting when adding individual probabilities:\[\operatorname{Pr}(A \cup B \cup C) = \operatorname{Pr}(A) + \operatorname{Pr}(B) + \operatorname{Pr}(C) - \operatorname{Pr}(A \cap B) - \operatorname{Pr}(A \cap C) - \operatorname{Pr}(B \cap C) + \operatorname{Pr}(A \cap B \cap C)\]This formula ensures that interactions among events are accounted for in their joint probabilities, avoiding double-counting areas where events overlap. Practicing with unions of events sharpens understanding of how events interact in probability.

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Most popular questions from this chapter

A computer dealer finds that the number of laptop computers her dealership sells each day is a random variable \(X\) where the probability distribution for \(X\) is given by $$ \operatorname{Pr}(X=x)= \begin{cases}\frac{c x^{2}}{x !}, & x=1,2,3,4,5 \\ 0, & \text { otherwise }\end{cases} $$ where \(c\) is a constant. Determine (a) the value of \(c\); (b) \(\operatorname{Pr}(X \geq 3)\); (c) \(\operatorname{Pr}(X=4 \mid X \geq 3)\); (d) \(E(X)\); and (e) \(\operatorname{Var}(X)\).

A carton contains 20 computer chips, four of which are defective. Isaac tests these chips \(-\) one at a time and without replacement - until he either finds a defective chip or has tested three chips. If the random variable \(X\) counts the number of chips Isaac tests, find (a) the probability distribution for \(X\); (b) \(\operatorname{Pr}(X \leq 2)\); (c) \(\operatorname{Pr}(X=1 \mid X \leq 2)\); (d) \(E(X)\); and (e) \(\operatorname{Var}(X)\).

Suppose that the number of boxes of cereal packaged each day at a certain packaging plant is a random variable - call it \(X-\) with \(E(X)=20,000\) boxes and \(\operatorname{Var}(X)=40,000\) boxes \(^{2}\). Use Chebyshev's Inequality to find a lower bound on the probability that the plant will package between 19,000 and 21,000 boxes of cereal on a particular day.

Let \(A, B \subseteq \mathbf{R}\), where \(A=\left\\{x \mid x^{2}-7 x \leq-12\right\\}\) and \(B=\) \(\left\\{x \mid x^{2}-x \leq 6\right\\}\). Determine \(A \cup B\) and \(A \cap B\).

Let \(\mathscr{S}\) be the sample space for an experiment \(\mathscr{E}\) and let \(A, B\) be events from \(\mathscr{P}\), where \(\operatorname{Pr}(A)=0.4, \operatorname{Pr}(B)=\) \(0.3\), and \(\operatorname{Pr}(A \cap B)=0.2\). Determine \(\operatorname{Pr}(\bar{A}), \operatorname{Pr}(\bar{B})\), \(\operatorname{Pr}(A \cup B), \operatorname{Pr}(\overline{A \cup B}), \operatorname{Pr}(A \cap \bar{B}), \operatorname{Pr}(\bar{A} \cap B)\) \(\operatorname{Pr}(A \cup \bar{B})\), and \(\operatorname{Pr}(\bar{A} \cup B) .\)

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