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Let \(S=\left\\{s_{1}, s_{2}, s_{3}, s_{4}, s_{5}\right\\}\) be the sample space associated with an experiment having the following probability distribution: $$\begin{array}{llllll} \hline \text { Outcome } & s_{1} & s_{2} & s_{3} & s_{4} & s_{5} \\ \hline \text { Probability } & \frac{1}{14} & \frac{3}{14} & \frac{6}{14} & \frac{2}{14} & \frac{2}{14} \\ \hline \end{array}$$ Find the probability of the event: a. \(A=\left\\{s_{1}, s_{2}, s_{4}\right\\}\) b. \(B=\left\\{s_{1}, s_{5}\right\\}\) c. \(C=S\)

Short Answer

Expert verified
a. \(P(A) = \frac{6}{14}\) b. \(P(B) = \frac{3}{14}\) c. \(P(C) = 1\)

Step by step solution

01

Find the probability of event A

To find the probability of event A, which is \(A=\{s_1, s_2, s_4\}\), we'll sum up the probabilities of each of these outcomes: $$P(A) = P(s_1) + P(s_2) + P(s_4)$$ Now, substitute the given probabilities: $$P(A) = \frac{1}{14} + \frac{3}{14} + \frac{2}{14}$$
02

Simplify the sum

Adding the probabilities together: $$P(A) = \frac{1 + 3 + 2}{14} = \frac{6}{14}$$ Since fraction is already in lowest term, the probability of event A is: $$P(A) = \frac{6}{14}$$
03

Find the probability of event B

To find the probability of event B, which is \(B=\{s_1, s_5\}\), we'll sum up the probabilities of each of these outcomes: $$P(B) = P(s_1) + P(s_5)$$ Now, substitute the given probabilities: $$P(B) = \frac{1}{14} + \frac{2}{14}$$
04

Simplify the sum

Adding the probabilities together: $$P(B) = \frac{1 + 2}{14} = \frac{3}{14}$$ Since fraction is already in lowest term, the probability of event B is: $$P(B) = \frac{3}{14}$$
05

Find the probability of event C

To find the probability of event C, which is \(C=S\), we'll sum up the probabilities of each of the outcomes: $$P(C) = P(s_1) + P(s_2) + P(s_3) + P(s_4) + P(s_5)$$ Now, substitute the given probabilities: $$P(C) = \frac{1}{14} + \frac{3}{14} + \frac{6}{14} + \frac{2}{14} + \frac{2}{14}$$
06

Simplify the sum

Adding the probabilities together: $$P(C) = \frac{1 + 3 + 6 + 2 + 2}{14} = \frac{14}{14} = 1$$ Since the event C contains all possible outcomes, its probability is equal to 1. The probabilities for the events A, B, and C are: a. \(P(A) = \frac{6}{14}\) b. \(P(B) = \frac{3}{14}\) c. \(P(C) = 1\)

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Key Concepts

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

Understanding Sample Space
The concept of a sample space is fundamental in probability theory. It represents the complete set of all possible outcomes of a particular experiment. If you think about conducting an experiment, the sample space composes all the results you might get.
In our given exercise, the sample space is denoted by the set \( S=\{s_1, s_2, s_3, s_4, s_5\} \). This means there are five possible outcomes when the experiment is conducted, and they are labeled as \( s_1, s_2, s_3, s_4, \) and \( s_5 \).
This is crucial because it helps set the boundaries for what the probability distribution will need to encompass.
What is a Probability Distribution?
A probability distribution assigns a probability to each outcome in the sample space. This tells us how likely each result is when an experiment is carried out. The sum of all assigned probabilities should equal 1, demonstrating that they account for the totality of outcomes.
For instance, in the exercise provided, the probability distribution is shown across the outcomes \( s_1 \) through \( s_5 \):
  • \( P(s_1) = \frac{1}{14} \)
  • \( P(s_2) = \frac{3}{14} \)
  • \( P(s_3) = \frac{6}{14} \)
  • \( P(s_4) = \frac{2}{14} \)
  • \( P(s_5) = \frac{2}{14} \)
Each of these probabilities represent the relative likelihood of their respective outcomes occurring. Summing them, we have, \( \frac{1}{14} + \frac{3}{14} + \frac{6}{14} + \frac{2}{14} + \frac{2}{14} = 1 \).
This confirms the validity of the probability distribution as a model for the sample space.
Calculating Event Probability
Event probability refers to the likelihood of a specific set of outcomes occurring. To find it, we simply add up the probabilities of all the individual outcomes that make up the event.
When solving for event probability, like with events \( A \), \( B \), and \( C \) from the exercise, you identify which outcomes form the event then sum their probabilities.
For event \( A \), consisting of outcomes \( \{s_1, s_2, s_4\} \), the probability is:
  • \( P(A) = P(s_1) + P(s_2) + P(s_4) = \frac{1}{14} + \frac{3}{14} + \frac{2}{14} = \frac{6}{14} \)
By doing this, you calculate how likely it is for the particular event (involving these specific outcomes) to occur when the experiment is performed.
This process is repeated similarly for events \( B \) and \( C \).
Identifying Outcomes in Probability
Outcomes in probability are the individual elements or results within a sample space after an experiment is conducted. Recognizing and understanding the implications of these outcomes help us build towards calculating probabilities.
In our exercise, each outcome is different (\( s_1, s_2, s_3, s_4, s_5 \)), and each can be part of different events.
If you imagine rolling a fair die, each face coming up (from 1 to 6) would be considered an outcome.
To solve cases involving probability, determine each possible outcome first. Then, you can start exploring and calculating associated probabilities by employing the defined sample space and its probability distribution.

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Most popular questions from this chapter

A study was conducted among a certain group of union members whose health insurance policies required second opinions prior to surgery. Of those members whose doctors advised them to have surgery, \(20 \%\) were informed by a second doctor that no surgery was needed. Of these, \(70 \%\) took the second doctor's opinion and did not go through with the surgery. Of the members who were advised to have surgery by both doctors, \(95 \%\) went through with the surgery. What is the probability that a union member who had surgery was advised to do so by a second doctor?

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 \(E\) is an event of an experiment, then \(P(E)+P\left(E^{c}\right)=1\).

A medical test has been designed to detect the presence of a certain disease. Among those who have the disease, the probability that the disease will be detected by the test is \(.95\). However, the probability that the test will erroneously indicate the presence of the disease in those who do not actually have it is .04. It is estimated that \(4 \%\) of the population who take this test have the disease. a. If the test administered to an individual is positive, what is the probability that the person actually has the disease? b. If an individual takes the test twice and both times the test is positive, what is the probability that the person actually has the disease? (Assume that the tests are independent.)

An experiment consists of randomly selecting one of three coins, tossing it, and observing the outcome-heads or tails. The first coin is a two-headed coin, the second is a biased coin such that \(P(\mathrm{H})=.75\), and the third is a fair coin. a. What is the probability that the coin that is tossed will show heads? b. If the coin selected shows heads, what is the probability that this coin is the fair coin?

Let \(E\) be any cvent in a sample space \(S .\) a. Are \(E\) and \(S\) independent? Explain your answer. b. Are \(E\) and \(\varnothing\) independent? Explain your answer.

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