/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 When an experiment is conducted,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When an experiment is conducted, one and only one of three mutually exclusive events \(S_{1}, S_{2}\) and \(S_{3}\), can occur, with \(P\left(S_{1}\right)=.2, P\left(S_{2}\right)=.5,\) and \(P\left(S_{3}\right)=.3 .\) The probabilities that an event A occurs, given that event \(S_{1}, S_{2}\), or \(S_{3}\) has occurred are $$ P\left(A \mid S_{1}\right)=.2 \quad P\left(A \mid S_{2}\right)=.1 \quad P\left(A \mid S_{3}\right)=.3 $$ If event A is observed, use this information to find the probabilities in Exercises 4 -6. \(P\left(S_{3} \mid A\right)\)

Short Answer

Expert verified
Answer: The probability is 0.5 or 50%.

Step by step solution

01

Write down the Bayes' theorem

Bayes' theorem relates the conditional probabilities of two events, say A and B. It is given by the formula: $$P(B | A) = \frac{P(A | B) * P(B)}{P(A)}$$ In our case, we want to find \(P(S_3 | A)\), so we will use the above formula with B = \(S_3\) and A = A.
02

Find the probability of event A

We are given the individual probabilities of events \(S_1\), \(S_2\), and \(S_3\) and the conditional probabilities \(P(A | S_1)\), \(P(A | S_2)\), and \(P(A | S_3)\). We need to find the overall probability of event A occurring. This can be done using the law of total probability. Since events \(S_1\), \(S_2\), and \(S_3\) are mutually exclusive and exhaustive, we can write: $$P(A) = P(A | S_1) * P(S_1) + P(A | S_2) * P(S_2) + P(A | S_3) * P(S_3)$$ Substituting the given values in the formula: $$P(A) = 0.2 * 0.2 + 0.1 * 0.5 + 0.3 * 0.3 = 0.04 + 0.05 + 0.09 = 0.18$$
03

Calculate the probability using the Bayes' theorem

Now we apply the Bayes' theorem to find \(P(S_3 | A)\). Plug in the numbers from step 1 and step 2 into the formula: $$P(S_3 | A) = \frac{P(A | S_3) * P(S_3)}{P(A)}$$ $$P(S_3 | A) = \frac{0.3 * 0.3}{0.18} = \frac{0.09}{0.18} = 0.5$$
04

Interpret the result

The probability of event \(S_3\) occurring, given that event A has occurred, is 0.5. This means, if we observe event A, there is a 50% chance that event \(S_3\) has occurred.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Bayes' theorem
Bayes' theorem is a way of finding a probability when we know certain other probabilities. The classic equation looks like this:
\[P(B | A) = \frac{P(A | B) \times P(B)}{P(A)}\]
What the theorem does is relate the conditional probability of event \(A\) happening given that \(B\) has occurred with the conditional probability of \(B\) happening given that \(A\) has occurred. This is incredibly useful in various fields, from medicine to machine learning, because it allows us to update our beliefs based on new evidence or data.
In the textbook exercise, we used Bayes' theorem to determine the probability of one of the mutually exclusive events \(S_3\) given that event \(A\) has occurred. The solution clearly lays out the process: first writing down Bayes' theorem, then finding the probability of event \(A\) using other provided probabilities, and finally using these to calculate \(P(S_3 | A)\). Remember, \(P(B)\) is the prior probability of \(B\), while \(P(B | A)\) is the likelihood and \(P(A)\) is evidence that needs to be calculated, often using the law of total probability.
Law of total probability
The law of total probability is crucial when dealing with scenarios involving multiple events that cover all possible outcomes, known as a partition of the sample space. When events are mutually exclusive (meaning they cannot happen at the same time) and exhaustive (they cover all possible outcomes), we can use this law to find the overall probability of a different event. The general formula looks like this:
\[P(A) = \sum P(A | B_i) \times P(B_i)\]
Here \(B_i\) represents each possible event. In the exercise, we knew the probability of event \(A\) given each mutually exclusive event \(S_1\), \(S_2\), and \(S_3\), and we also knew the probabilities of each of these events. By multiplying the probability of \(A\) happening given each separate event by the probability of the event itself, and then adding them up, we obtained the total probability of event \(A\), which was used later on in Bayes' theorem.
Mutually exclusive events
When we talk about mutually exclusive events, we're discussing scenarios where the occurrence of one event means that none of the other events can happen at the same time. For example, flipping a coin can only result in either heads or tails—these are mutually exclusive outcomes. In probability theory, this means that the intersection of these events (where both happen at once) has a probability of zero. Mathematically, this is written as:
\[P(A \cap B) = 0\]
In our textbook problem, \(S_1\), \(S_2\), and \(S_3\) are mutually exclusive events, and the sum of their probabilities adds up to 1, which is a characteristic of such events forming a complete set. Understanding this concept is important for correctly applying the law of total probability, as it assures us that we are accounting for every possible outcome once, and only once, making sure our computed probability for event \(A\) is accurate and meaningful.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two cold tablets are unintentionally put in a box containing two aspirin tablets, that appear to be identical. One tablet is selected at random from the box and swallowed by the first patient. The second patient selects another tablet at random and swallows it. a. List the simple events in the sample space \(S\). b. Find the probability of event \(A\), that the first patient swallowed a cold tablet. c. Find the probability of event \(B\), that exactly one of the two patients swallowed a cold tablet. d. Find the probability of event \(C,\) that neither patient swallowed a cold tablet.

A single person is hired to taste and rank each of three brands of tea, which are unmarked except for identifying symbols \(A, B,\) and \(C .\) If the taster has no ability to distinguish a difference in taste among teas, what is the probability that the taster will rank tea type \(A\) as the most desirable? As the least desirable?

Experiment II A sample space contains seven simple events: \(E_{1}, E_{2}, \ldots, E_{7} .\) Suppose that \(E_{1}, E_{2}, \ldots, E_{6}\) all have the same probability, but \(E_{7}\) is twice as likely as the others. Find the probabilities of the events. $$ C=\left\\{E_{2}, E_{4}\right\\} $$

Suppose \(P(A)=.1\) and \(P(B)=.5 .\) $$\text { If } P(A \mid B)=.1, \text { are } A \text { and } B \text { independent? }$$

Suppose that a certain disease is present in \(10 \%\) of the population, and that there is a screening test designed to detect this disease if present. The test does not always work perfectly. Sometimes the test is negative when the disease is present, and sometimes it is positive when the disease is absent. The following table shows the proportion of times that the test produces various results. $$ \begin{array}{lcc} \hline & \text { Test Is Positive }(P) & \text { Test Is Negative }(N) \\ \hline \text { Disease } & .08 & .02 \\ \text { Present }(D) & & \\ \text { Disease } & .05 & .85 \\ \text { Absent }(D 9 & & \\ & & \\ \hline \end{array} $$ a. Find the following probabilities from the table: \(P(D), P\left(D^{c}\right), P\left(N \mid D^{c}\right), P(N \mid D)\) b. Use Bayes' Rule and the results of part a to find \(P(D \mid N)\) c. Use the definition of conditional probability to find \(P(D \mid N)\). (Your answer should be the same as the answer to part b.) d. Find the probability of a false positive, that the test is positive, given that the person is disease-free. e. Find the probability of a false negative, that the test is negative, given that the person has the disease. f. Are either of the probabilities in parts d or e large enough that you would be concerned about the reliability of this screening method? Explain.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.