/*! 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 70 If an experiment is conducted, o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If an experiment is conducted, one and only one of three mutually exclusive events \(S_{1}, S_{2},\) and \(S_{3}\) can occur, with these probabilities: $$P\left(S_{1}\right)=.2 \quad P\left(S_{2}\right)=.5 \quad P\left(S_{3}\right)=.3$$ The probabilities of a fourth event \(A\) occurring, given that event \(S_{1}, S_{2},\) or \(S_{3}\) occurs, 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, find \(P\left(S_{1} \mid A\right), P\left(S_{2} \mid A\right),\) and \(P\left(S_{3} \mid A\right)\).

Short Answer

Expert verified
Question: Given the probabilities $P(S_1) = 0.2, P(S_2) = 0.5, P(S_3) = 0.3$, and $P(A|S_1) = 0.2, P(A|S_2) = 0.1, P(A|S_3) = 0.3$, find the conditional probabilities $P(S_1|A)$, $P(S_2|A)$, and $P(S_3|A)$. Answer: The conditional probabilities are $P(S_1|A) = 0.25$, $P(S_2|A) = 0.3125$, and $P(S_3|A) = 0.5625$.

Step by step solution

01

Recall Bayes' theorem

Bayes' theorem states that for events \(B\) and \(C\): $$P(B|C) = \frac{P(C|B) P(B)}{P(C)}$$ In this case, we need to find \(P(S_{1} \mid A)\), \(P(S_{2} \mid A)\), and \(P(S_{3} \mid A)\), where \(B\) is one of the events \(S_1, S_2, S_3\) and \(C\) is the event \(A\).
02

Express the probabilities

From the given information, we know: $$P(S_{1}) = 0.2, P(S_{2}) = 0.5, P(S_{3}) = 0.3$$ $$P(A|S_{1}) = 0.2, P(A|S_{2}) = 0.1, P(A|S_{3}) = 0.3$$
03

Calculate the probability of A

We need to compute \(P(A)\), which can be obtained using the law of total probability: $$P(A) = P(A|S_{1})P(S_{1}) + P(A|S_{2})P(S_{2}) + P(A|S_{3})P(S_{3})$$ Plugging the values we know: $$P(A) = 0.2 \cdot 0.2 + 0.1 \cdot 0.5 + 0.3 \cdot 0.3 = 0.16$$
04

Calculate the conditional probabilities

Now we can use Bayes' theorem to find the probabilities we need: 1. For \(P(S_{1} \mid A)\): $$P(S_{1}|A) = \frac{P(A|S_{1})P(S_{1})}{P(A)} = \frac{0.2 \cdot 0.2}{0.16} = \frac{1}{4} = 0.25$$ 2. For \(P(S_{2} \mid A)\): $$P(S_{2}|A) = \frac{P(A|S_{2})P(S_{2})}{P(A)} = \frac{0.1 \cdot 0.5}{0.16} = \frac{5}{16} = 0.3125$$ 3. For \(P(S_{3} \mid A)\): $$P(S_{3}|A) = \frac{P(A|S_{3})P(S_{3})}{P(A)} = \frac{0.3 \cdot 0.3}{0.16} = \frac{9}{16} = 0.5625$$ So we have found: $$P(S_{1} \mid A) = 0.25, P(S_{2} \mid A) = 0.3125, P(S_{3} \mid A) = 0.5625$$

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.

Conditional Probability
Conditional probability is a fundamental concept in probability theory. It refers to the probability of an event occurring given that another event has already happened. In the context of the given exercise, we are interested in the probabilities of events \(S_1\), \(S_2\), and \(S_3\) happening given that event \(A\) has occurred. This is denoted as \(P(S_i | A)\).

To calculate these probabilities, we use Bayes' Theorem, which helps in updating our probability estimates based on new information. Using Bayes' Theorem, the conditional probability \(P(S_i | A)\) is calculated using the formula:
  • \(P(S_i | A) = \frac{P(A|S_i) P(S_i)}{P(A)}\)
Here, \(P(A | S_i)\) is the probability of \(A\) occurring given \(S_i\), \(P(S_i)\) is the initial probability of \(S_i\), and \(P(A)\) is the probability of \(A\) occurring at all. This approach helps to refine probability assessments when additional conditions are known.
Law of Total Probability
The Law of Total Probability is a crucial principle used to determine the probability of an event by considering all possible ways the event can occur. This law states that if \(S_1, S_2,\) and \(S_3\) are exhaustive and mutually exclusive events, the probability of any event \(A\) can be calculated by considering the conditional probabilities with these events.

The formula is:
  • \(P(A) = P(A|S_{1})P(S_{1}) + P(A|S_{2})P(S_{2}) + P(A|S_{3})P(S_{3})\)
This formula breaks down the probability of \(A\) into parts based on \(A\)'s occurrence in connection with each \(S_i\).

In the exercise, it helps us calculate \(P(A)\) by adding up the weighted probabilities of \(A\) under different scenarios, where the weights are the probabilities of \(S_i\). Knowing \(P(A)\) is essential for applying Bayes' Theorem, as it serves as the denominator when calculating any conditional probability \(P(S_i|A)\).
Mutually Exclusive Events
Mutually exclusive events are events that cannot happen at the same time. In probability terms, if \(S_1, S_2, \text{and } S_3\) are mutually exclusive, the occurrence of one prevents any others from occurring simultaneously. This means \(P(S_i \cap S_j) = 0\) for any distinct events \(S_i\) and \(S_j\).

In the exercise, \(S_1, S_2,\) and \(S_3\) are mutually exclusive events, which is why the sum of their probabilities equals 1:
  • \(P(S_1) + P(S_2) + P(S_3) = 1\)
This property is used to ensure that one and only one of these events will occur, simplifying the calculation of \(P(A)\) using the Law of Total Probability. Understanding mutually exclusive events is essential because it narrows down possible outcomes to non-overlapping probabilities, allowing for straightforward computations when analyzing probability scenarios such as this one.

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

You have \(t\) wo groups of distinctly different items, 10 in the first group and 8 in the second. If you select one item from each group, how many different pairs can you form?

A random variable \(x\) has this probability distribution: $$\begin{array}{l|llllll}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .1 & .3 & .4 & .1 & ? & .05\end{array}$$ a. Find \(p(4)\). b. Construct a probability histogram to describe \(p(x)\). c. Find \(\mu, \sigma^{2},\) and \(\sigma\). d. Locate the interval \(\mu \pm 2 \sigma\) on the \(x\) -axis of the histogram. What is the probability that \(x\) will fall into this interval? e. If you were to select a very large number of values of \(x\) from the population, would most fall into the interval \(\mu \pm 2 \sigma\) ? Explain.

A certain virus afflicted the families in three adjacent houses in a row of 12 houses. If houses were randomly chosen from a row of 12 houses, what is the probability that the three houses would be adjacent? Is there reason to believe that this virus is contagious?

You can insure a \(\$ 50,000\) diamond for its total value by paying a premium of \(D\) dollars. If the probability of loss in a given year is estimated to be .01 , what premium should the insurance company charge if it wants the expected gain to equal \$1000?

Most coffee drinkers take a little time each day for their favorite beverage, and many take more than one coffee break every day. The table below, adapted from a Snapshot in \(U S A\) Today, shows the probability distribution for \(x\), the number of daily coffee breaks taken per day by coffee drinkers. \({ }^{8}\) $$\begin{array}{l|lllllll}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .28 & .37 & .17 & .12 & .05 & .01\end{array}$$ a. What is the probability that a randomly selected coffee drinker would take no coffee breaks during the day? b. What is the probability that a randomly selected coffee drinker would take more than two coffee breaks during the day? c. Calculate the mean and standard deviation for the random variable \(x\). d. Find the probability that \(x\) falls into the interval \(\mu \pm 2 \sigma\).

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.