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Consider a sample space \(\Omega=\\{a, e, i, o, u\\}\) endowed with the following probability density: \(p(a)=0.22, p(e)=0.35, p(i)=0.13, p(o)=0.20,\) and \(p(u)=0.10 .\) Determine the probabilities of the following events: (a) \(\\{a, o\\}\) (b) \(\emptyset\) (c) The event \(E\) consisting of all those outcomes in \(\Omega\) that come after the letter \(k\) in the alphabet

Short Answer

Expert verified
(a) 0.42; (b) 0; (c) 0.30.

Step by step solution

01

Understand the Given Data

We are given a sample space \(\Omega = \{a, e, i, o, u\}\) with the probabilities for each element: \(p(a) = 0.22\), \(p(e) = 0.35\), \(p(i) = 0.13\), \(p(o) = 0.20\), and \(p(u) = 0.10\). Our task is to find the probability of different events described using this sample space.
02

Calculate Probability of Event \(\{a, o\}\)

Event \(\{a, o\}\) consists of the outcomes \(a\) and \(o\). The probability of an event is the sum of the probabilities of the individual outcomes in the event. Thus: \[P(\{a, o\}) = p(a) + p(o) = 0.22 + 0.20 = 0.42.\]
03

Calculate Probability of the Empty Set \(\emptyset\)

The empty set \(\emptyset\) contains no outcomes. By definition, the probability of the empty set is always 0: \[P(\emptyset) = 0.\]
04

Define and Calculate Probability for Event \(E\)

Event \(E\) includes all outcomes in \(\Omega\) that come after the letter \(k\). In the alphabet, the letters \(o\) and \(u\) come after \(k\), so \(E = \{o, u\}\). Calculate the probability: \[P(E) = p(o) + p(u) = 0.20 + 0.10 = 0.30.\]

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Key Concepts

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

Probability Density
Probability density is an essential concept in understanding the likelihood of outcomes in a sample space. In this context, a sample space is a set of all possible outcomes of an experiment or situation. Here, our sample space is \(\Omega = \{a, e, i, o, u\}\), each with its own probability.
Probability density is a function that assigns a probability to each outcome in the sample space. In the given example, probabilities are provided for each element:
  • \(p(a) = 0.22\)
  • \(p(e) = 0.35\)
  • \(p(i) = 0.13\)
  • \(p(o) = 0.20\)
  • \(p(u) = 0.10\)
The sum of these probabilities must equal 1, ensuring that they map out the entire sample space. Understanding how to assign and manipulate these probabilities is crucial in calculating the probability of various events.
Events
In probability theory, an event refers to a subset of outcomes within the sample space. Events can be simple or compound, and their probability is calculated by summing up the probabilities of their respective outcomes.
For instance, we determined the probability of event \(\{a, o\}\), which includes the outcomes 'a' and 'o'. Thus, the probability of the event is computed as:\[P(\{a, o\}) = p(a) + p(o) = 0.22 + 0.20 = 0.42.\]Events can vary broadly and can include specific features, such as containing particular letters or conditions.
In solving problems, clearly defining the events helps simplify the calculation by breaking down the sample space into manageable parts.
Alphabetical Order
Alphabetical order often aids in determining conditions within a sample space. It is a straightforward method where items are arranged in alphabetical sequence, which can be helpful for understanding positions or sequences of characters or events.
For our exercise, part of determining probabilities involved distinguishing outcomes that come after a particular letter, \'k\', alphabetically within the set \(\Omega = \{a, e, i, o, u\}\).
The event described as outcomes coming after \'k\' resulted in \'o\' and \'u\'. Understanding this sequence allowed us to form event \(E\), where \(E = \{o, u\}\), making it easy to add up the respective probabilities to find:\[P(E) = p(o) + p(u) = 0.20 + 0.10 = 0.30.\]Alphabetical order is not just for arranging characters but becomes a useful tool in sorting and evaluating conditions on data sets.
Empty Set
The empty set, denoted as \(\emptyset\), is a fundamental notion in set theory. It contains no elements and is a unique set with a probability of zero in probability theory.
Calculating the probability of the empty set is always straightforward, as it does not represent any outcomes or scenarios within the sample space.
By definition,\[P(\emptyset) = 0.\]This underscores that without any possible events occurring, there's no likelihood of such a scenario unfolding.
In practice, the concept of the empty set is useful for representing scenarios or conditions where no outcomes meet specific criteria, ensuring clarity and precision in probability calculations.

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Most popular questions from this chapter

Suppose \(A\) and \(B\) are events in a sample space such that \(P(A)=1 / 4, P(B)=5 / 8\). and \(P(A \cup B)=3 / 4 .\) What is \(P(A \cap B) ?\)

Two manufacturing companies \(M_{1}\) and \(M_{2}\) produce a certain unit that is used in an assembly plant. Company \(M_{1}\) is larger than \(M_{2}\), and it supplies the plant with twice as many units per day as \(M_{2}\) does. \(M_{1}\) also produces more defects than \(M_{2}\). Because of past experience with these suppliers, it is felt that \(10 \%\) of \(M_{1}\) 's units have some defect, whereas only \(5 \%\) of \(M_{2}\) 's units are defective. Now, suppose that a unit is selected at random from a bin in the assembly plant. (a) What is the probability that the unit was supplied by company \(M_{1} ?\) (b) What is the probability that the unit is defective? (c) What is the probability that the unit was supplied by \(M_{1}\) if the unit is defective?

The probability density function for the random variable \(X\) defined to be the number of cars owned by a randomly selected family in Millinocket is given as $$\begin{array}{l|c|c|c|c|c}x & 0 & 1 & 2 & 3 & 4 \\\\\hline p(X=x) & 0.08 & 0.15 & 0.45 & 0.27 & 0.05\end{array}$$ Compute the variance and standard deviation of \(X\).

In a fierce battle, not less than \(70 \%\) of the soldiers lost one eye, not less than \(75 \%\) lost one ear, not less than \(80 \%\) lost one hand, and not less than \(85 \%\) lost one leg. What is the smallest percentage who could have lost simultaneously one car, one eye, one hand, and one leg? This problem comes from Tangled Tales by Lewis Carroll, the author of Alice in Wonderland.

Suppose that sample space \(\Omega_{1}\) is chosen to model the experiment of rolling a pair of dice and that the probability density function \(p\) assigned to \(\Omega_{1}\) is \(p(\omega)=1 / 36\) for \(\omega \in \Omega_{1}\). Under these assumptions, compute the probability of rolling a sum of 3. Compare your answer to the answers of \(1 / 18\) and \(1 / 11\) obtained in the text, and discuss.

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