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Let \(S=\left\\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, s_{6}\right\\}\) be the sample space associated with an experiment having the probability distribution shown in the accompanying table. If \(A=\left\\{s_{1}, s_{2}\right\\}\) and \(B=\left\\{s_{1}, s_{5}, s_{6}\right\\}\), find a. \(P(A), P(B)\) b. \(P\left(A^{c}\right), P\left(B^{c}\right)\) c. \(P(A \cap B)\) d. \(P(A \cup B)\) e. \(P\left(A^{c} \cap B^{c}\right)\) f. \(P\left(A^{c} \cup B^{c}\right)\) $$ \begin{array}{cc} \hline \text { Outcome } & \text { Probability } \\ \hline s_{1} & \frac{1}{3} \\ \hline s_{2} & \frac{1}{8} \\ \hline s_{3} & \frac{1}{6} \\ \hline s_{4} & \frac{1}{6} \\ \hline s_{5} & \frac{1}{12} \\ \hline s_{6} & \frac{1}{8} \\ \hline \end{array} $$

Short Answer

Expert verified
a. \(P(A) = \dfrac{5}{8}\) and \(P(B) = \dfrac{7}{12}\) b. \(P\left(A^{c}\right) = \dfrac{3}{8}\) and \(P\left(B^{c}\right) = \dfrac{5}{12}\) c. \(P(A \cap B) = \dfrac{1}{3}\) d. \(P(A \cup B) = \dfrac{11}{12}\) e. \(P\left(A^{c} \cap B^{c}\right) = \dfrac{1}{3}\) f. \(P\left(A^{c} \cup B^{c}\right) = \dfrac{1}{12}\)

Step by step solution

01

a. Finding \(P(A)\) and \(P(B)\)

To find the probability of events A and B, we simply need to add the probabilities of their elements. Event A = {s1, s2}: \(P(A) = P(s_{1}) + P(s_{2}) = \dfrac{1}{3} + \dfrac{1}{8} = \dfrac{5}{8}\) Event B = {s1, s5, s6}: \(P(B) = P(s_{1}) + P(s_{5}) + P(s_{6}) = \dfrac{1}{3} + \dfrac{1}{12} + \dfrac{1}{8} = \dfrac{7}{12}\)
02

b. Finding \(P(A^{c})\) and \(P(B^{c})\)

Now we will find the probabilities for the complements of events A and B by subtracting the probability of each event from 1: \(P(A^{c}) = 1 - P(A) = 1 - \dfrac{5}{8} = \dfrac{3}{8}\) \(P(B^{c}) = 1 - P(B) = 1 - \dfrac{7}{12} = \dfrac{5}{12}\)
03

c. Finding \(P(A \cap B)\)

Next, we find the probability of the intersection of events A and B, which is the common outcome(s) between both events: A ∩ B = {s1}: \(P(A \cap B) = P(s_{1}) = \dfrac{1}{3}\)
04

d. Finding \(P(A \cup B)\)

Now, let's find the probability of the union of events A and B: \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = \dfrac{5}{8} + \dfrac{7}{12} - \dfrac{1}{3} = \dfrac{11}{12}\)
05

e. Finding \(P(A^{c} \cap B^{c})\)

To find the probability of the intersection of the complements of events A and B, we can first identify the elements in both complements: \(A^{c} = \{s_{3}, s_{4}, s_{5}, s_{6}\}\) \(B^{c} = \{s_{2}, s_{3}, s_{4}\}\) \(A^{c} \cap B^{c} = \{s_{3}, s_{4}\}\) \(P(A^{c} \cap B^{c}) = P(s_{3}) + P(s_{4}) = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{3}\)
06

f. Finding \(P(A^{c} \cup B^{c})\)

Finally, let's find the probability of the union of the complements of events A and B: \(P(A^{c} \cup B^{c}) = P(A^{c}) + P(B^{c}) - P(A^{c} \cap B^{c}) = \dfrac{3}{8} + \dfrac{5}{12} - \dfrac{1}{3} = \dfrac{1}{12}\)

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Key Concepts

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

Combinatorics
Combinatorics is the branch of mathematics dealing with counting, arrangement, and combination of objects. In probability, combinatorics helps us determine the number of ways an event can occur, which is crucial for calculating probabilistic outcomes.
When dealing with sets and probability, combinatorics tells us how many subsets or arrangements can be formed. For instance, when you have a sample space, calculating the probabilities of different events often relies on understanding combinations and permutations.
In this specific exercise, combinatorics helps us identify the elements within sets A and B, and their respective intersections and unions. This approach aids in laying the groundwork for subsequent probability calculations, showing the importance of effectively counting the arrangements or combinations involved.
Event Probability
Event probability is the measure of how likely an event is to occur, expressed as a value between 0 (impossible event) and 1 (certain event). To find this, divide the number of favorable outcomes by the total number of possible outcomes in the sample space.
In the exercise, we calculated the event probabilities for A and B by summing the probabilities of individual outcomes in each event. This approach gives us the likelihood of event A or B occurring by adding the associated probabilities:
  • Event A: {s1, s2}, with probabilities for each element added together.
  • Event B: {s1, s5, s6}, similarly computed by summing the probabilities of these particular outcomes.
Understanding event probability allows you to predict how often an event will occur in repeated trials of an experiment.
Sample Space
The sample space is the complete set of all possible outcomes of an experiment. In probability theory, it's crucial to precisely define this space to effectively analyze any event.
For the exercise, the sample space is defined as \(S=\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}, s_{6}\}\). Each outcome has an associated probability, summing to 1, as it covers all possible results. By considering different subsets of outcomes, we can calculate the probabilities of more complex events, like unions or intersections of sets.
The sample space acts as the foundational structure upon which we build further probability concepts, ensuring that all potential results are considered when calculating probabilities.
Probability Distribution
A probability distribution assigns probabilities to each outcome in the sample space. These probabilities indicate how likely each outcome is to occur within the context of an experiment.
The table from the exercise presents a probability distribution over the sample space \(S\), with each outcome \(s_{i}\) having a specific probability:
  • \(s_{1}: \frac{1}{3}\)
  • \(s_{2}: \frac{1}{8}\)
  • \(s_{3}: \frac{1}{6}\)
  • \(s_{4}: \frac{1}{6}\)
  • \(s_{5}: \frac{1}{12}\)
  • \(s_{6}: \frac{1}{8}\)
It's important to note that the probabilities in the distribution must sum to 1, confirming the completeness of the sample space. Understanding the probability distribution allows us to calculate event probabilities more efficiently, as it provides the likelihood of each basic outcome in a structured format.

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Most popular questions from this chapter

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. If \(A\) is a subset of \(B\), then \(P(A) \leq P(B)\).

Human blood is classified by the presence or absence of three main antigens (A, B, and Rh). When a blood specimen is typed, the presence of the \(\mathrm{A}\) and/or \(\mathrm{B}\) antigen is indicated by listing the letter \(A\) and/or the letter \(B\). If neither the A nor B antigen is present, the letter \(\mathrm{O}\) is used. The presence or absence of the \(\mathrm{Rh}\) antigen is indicated by the symbols \(+\) or \(-\), respectively. Thus, if a blood specimen is classified as \(\mathrm{AB}^{+}\), it contains the \(\mathrm{A}\) and the \(\mathrm{B}\) antigens as well as the \(\mathrm{Rh}\) antigen. Similarly, \(\mathrm{O}^{-}\) blood contains none of the three antigens. Using this information, determine the sample space corresponding to the different blood groups.

The following table gives the number of people killed in rollover crashes in various types of vehicles in 2002 : Find the empirical probability distribution associated with these data. If a fatality due to a rollover crash in 2002 is picked at random, what is the probability that the victim was in a. \(\mathrm{A}\) car? b. An SUV? c. A pickup or an SUV?

The results of a recent television survey of American TV households revealed that 87 out of every 100 TV households have at least one remote control. What is the probability that a randomly selected TV household does not have at least one remote control?

Explain why the statement is incorrect. A red die and a green die are tossed. The probability that a 6 will appear uppermost on the red die is \(\frac{1}{6}\), and the probability that a 1 will appear uppermost on the green die is \(\frac{1}{6}\). Hence, the probability that the red die will show a 6 or the green die will show a 1 is \(\frac{1}{6}+\frac{1}{6}\).

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