/*! 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 1 In a sample space \(S\), we have... [FREE SOLUTION] | 91Ó°ÊÓ

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In a sample space \(S\), we have the following events and some associated probabilities: $$P(E)=\frac{2}{3}, P(A | E)=\frac{3}{50} \text { and } P\left(A | E^{\prime}\right)=\frac{1}{25}$$ a) Represent the information using a tree diagram. b) Find \(P(A)\) c) Find \(P(E | A)\)

Short Answer

Expert verified
\(P(A) = \frac{4}{75}\), \(P(E|A) = \frac{3}{4}\).

Step by step solution

01

Understand Given Events and Probabilities

We have a sample space \(S\) with events \(E\), \(A\), and their complements. We are given the probabilities \(P(E)=\frac{2}{3}\), \(P(A|E)=\frac{3}{50}\), and \(P(A|E')=\frac{1}{25}\). We need to represent these in a tree diagram and find other probabilities.
02

Draw the Tree Diagram

A tree diagram consists of branches showing sequences of events. Start with the event \(E\) with probability \(\frac{2}{3}\) and its complement \(E'\) with probability \(\frac{1}{3}\). From \(E\), draw branches for \(A\) with \(P(A|E) = \frac{3}{50}\) and \(A'\) with \(P(A'|E) = 1 - \frac{3}{50}\). From \(E'\), draw branches for \(A\) with \(P(A|E') = \frac{1}{25}\) and \(A'\) with \(P(A'|E') = 1 - \frac{1}{25}\).
03

Multiply Along the Branches to Get Joint Probabilities

Calculate \(P(E \cap A) = P(E) \cdot P(A|E) = \frac{2}{3} \times \frac{3}{50} = \frac{1}{25}\). Also calculate \(P(E' \cap A) = P(E') \cdot P(A|E') = \frac{1}{3} \times \frac{1}{25} = \frac{1}{75}\).
04

Find Total Probability of A, \(P(A)\)

Use the law of total probability: \(P(A) = P(E \cap A) + P(E' \cap A) = \frac{1}{25} + \frac{1}{75}\). Convert to a common denominator: \(\frac{3}{75} + \frac{1}{75} = \frac{4}{75}\).
05

Calculate \(P(E|A)\) Using Bayes' Theorem

Apply Bayes' theorem: \(P(E|A) = \frac{P(E \cap A)}{P(A)} = \frac{\frac{1}{25}}{\frac{4}{75}}\). Simplify: \(\frac{1}{25} \div \frac{4}{75} = \frac{1}{25} \times \frac{75}{4} = \frac{3}{4}\).

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Key Concepts

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

Tree Diagram
A tree diagram is a great tool to visualize the sequence of events and their probabilities in a clear, organized manner. In the context of probability, each branch represents an event, while its associated probability is the likelihood of that event occurring. Let's consider the previous exercise, To represent the events and probabilities in a tree diagram, start with the initial event, which is event \(E\) occurring with a probability of \(\frac{2}{3}\). The complementary event \(E'\) occurs with a probability of \( \frac{1}{3} \).
  • From the event \(E\), branches will lead to event \(A\) with a conditional probability \(P(A|E) = \frac{3}{50}\) and its complement \(A'\) with probability \(1 - \frac{3}{50}\). This ensures that probabilities from the same event sum up to one.
  • Similarly, from event \(E'\), branches lead to \(A\) with probability \(P(A|E') = \frac{1}{25}\) and to \(A'\) with probability \(1 - \frac{1}{25}\).
This visual representation helps in intuitively understanding how different events are connected and how their probabilities are interdependent. The subsequent step is to compute the joint probabilities along these branches. Multiply the probabilities along the paths from the start event through each branch to the terminal event to find joint probabilities.
Bayes' Theorem
Bayes' Theorem provides a way to update the probability estimate for an event based on new evidence. It connects the original or prior probability of an event with its likelihood and the evidence's prior probability.Bayes' Theorem is expressed as:\[P(E|A) = \frac{P(A|E) \cdot P(E)}{P(A)}\]In our exercise, we used Bayes' theorem to find the probability of event \(E\) happening, given that \(A\) has occurred. By utilizing the values of conditional probabilities \(P(A|E)\) and \(P(A|E')\) and the prior probability \(P(E)\), we calculated the updated probability \(P(E|A)\).Solving the exercise, we had:
  • Given \(P(E \cap A) = \frac{1}{25}\) and \(P(A) = \frac{4}{75}\), the probability \(P(E|A)\) was calculated using Bayes' theorem:
  • \[ P(E|A) = \frac{\frac{1}{25}}{\frac{4}{75}} = \frac{3}{4} \]
Understanding Bayes' Theorem is essential as it allows for revising estimates as more information becomes available, providing a dynamic way to approach probability problems. Whether in medical testing, spam filtering, or complex scenarios, Bayes' theorem is the mathematical backbone for decision-making under uncertainty.
Law of Total Probability
The Law of Total Probability plays a vital role in finding the probability of a particular event by considering all possible ways that the event can occur. It involves breaking down the probability of the event into its constituent parts, using known conditional probabilities.In the context of the exercise, we were required to find the probability \(P(A)\), the likelihood of event \(A\) occurring, using the Law of Total Probability. The rule is calculated as follows:\[P(A) = P(E \cap A) + P(E' \cap A)\]Here, you multiply the probability of the intersecting events and sum these results. Utilizing the information, the law enabled us to calculate:
  • \(P(E \cap A) = \frac{2}{3} \times \frac{3}{50} = \frac{1}{25} \)
  • \(P(E' \cap A) = \frac{1}{3} \times \frac{1}{25} = \frac{1}{75}\)
After converting these to a common denominator, the total probability \(P(A)\) was:\[P(A) = \frac{1}{25} + \frac{1}{75} = \frac{4}{75}\]The Law of Total Probability is helpful in determining the probability of events that can occur through multiple, mutually exclusive pathways, thus providing a comprehensive probability assessment of the event.

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Most popular questions from this chapter

Two dice are rolled and the numbers on the top face are observed. a) List the elements of the sample space. b) Let \(x\) represent the sum of the numbers observed. Copy and complete the following table. $$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|} \hline x & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline P(x) & & \frac{1}{18} & & & & & & & & & \\ \hline \end{array}$$ c) What is the probability that at least one die shows a \(6 ?\) d) What is the probability that the sum is at most \(10 ?\) e) What is the probability that a die shows 4 or the sum is \(10 ?\) f) Given that the sum is \(10,\) what is the probability that one of the dice is a \(4 ?\)

A box contains three balls, blue, green and yellow. You run an experiment where you draw a ball, look at its color and then replace it and draw a second ball. a) What is the sample space of this experiment? b) What is the event of drawing yellow first? c) What is the event of drawing the same color twice?

If events \(A\) and \(B\) are given such that \(P(A)=\frac{1}{3}, P(A \cup B)=\frac{4}{9}\) and \(P(B)=\frac{2}{9},\) show that \(A\) and \(B\) are neither independent nor mutually exclusive.

A school has 250 employees categorized by task and gender in the following table. $$\begin{array}{|l|c|c|c|}\hline & \text { Teaching } & \text { Administrative } & \text { Support } \\\\\hline \text { Male } & 84 & 14 & 52 \\\\\hline \text { Female } & 56 & 26 & 18 \\\\\hline\end{array}$$ An employee is randomly selected. Let \(A\) be the event that he/she is an administrative staff member, T teaching staff, S support, \(M\) male, and F female. a) Write down the following probabilities: \(P(F), P(F \cap T), P\left(F \cup A^{\prime}\right), P\left(F^{\prime} | A\right)\) b) Which events are independent of \(F\), which are mutually exclusive to \(F\). Justify your choices. c) (i) Given that \(90 \%\) of teachers, as well as \(80 \%\) of the administrative staff and \(30 \%\) of the support staff, own cars, find the probability that a staff member chosen at random owns a car. (ii) Knowing that the randomly chosen staff member owns a car, find the probability that he/she is a teacher.

Five cards are chosen at random from a deck of 52 cards. Find the probability that the set contains a) 3 kings b) 4 hearts and 1 diamond.

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