/*! 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 2 Suppose \(P(X \mid Y)=1 / 3\) an... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose \(P(X \mid Y)=1 / 3\) and \(P(Y)=1 / 4\). What is \(P(X \cap Y) ?\)

Short Answer

Expert verified
The probability of the intersection of events \(X\) and \(Y\) is \(\frac{1}{12}\).

Step by step solution

01

Write the given probabilities

We know the following probabilities: - \(P(X \mid Y) = \frac{1}{3}\) - \(P(Y) = \frac{1}{4}\) We will use these probabilities to find \(P(X \cap Y)\).
02

Use the formula for conditional probability

Recall the formula for conditional probability: \[P(X \mid Y) = \frac{P(X \cap Y)}{P(Y)}\]
03

Rearrange the formula to find \(P(X \cap Y)\)

To find \(P(X \cap Y)\), we can rearrange the formula as follows: \[P(X \cap Y) = P(X \mid Y) \times P(Y)\]
04

Substitute the given probabilities into the formula

Now, plug in the given probabilities into the formula: \[P(X \cap Y) = \frac{1}{3} \times \frac{1}{4}\]
05

Calculate the result

Finally, multiply the fractions to get the probability of the intersection: \[P(X \cap Y) = \frac{1}{12}\] Thus, the probability of the intersection of events \(X\) and \(Y\) is \(\frac{1}{12}\).

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.

Probability of Intersection
Understanding the probability of intersection is crucial in determining how often two events will happen at the same time. This concept is used when we want to know the chance of two events, say event A and event B, both occurring. It's represented mathematically as \( P(A \cap B) \), where \( \cap \) signifies 'and', or the intersection between events A and B.

Let's take a moment to visualize: imagine you're flipping a coin and spinning a spinner. If we wanted to find the probability of flipping heads (event A) and the spinner landing on blue (event B), we would be seeking the probability of their intersection. To calculate this, we need to know the individual probabilities and how the events interact—are they independent, or does one affect the outcome of the other? In our education platform example, we found out that, based on the given conditional probability, \( P(X \cap Y) = \frac{1}{12} \), which provides us with the likelihood of both events X and Y happening together.
Bayes' Theorem
Named after Reverend Thomas Bayes, Bayes' Theorem is a cornerstone of probability theory, providing a way to update the probability of a hypothesis as more evidence becomes available. The theorem is written as:
\[ P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)} \]
Where \(P(A \mid B)\) is the probability of event A given event B has occurred, \(P(B \mid A)\) is the probability of event B given A has occurred, \(P(A)\) is the probability of event A, and \(P(B)\) is the probability of event B.

Bayes' theorem allows for the recalculation of probabilities when new data is given, which makes it an indispensable tool in many fields like statistics, finance, and even machine learning, as well as in everyday decision-making. For our example, if we had the reverse conditional probability \(P(Y \mid X)\), we could use Bayes' Theorem to update our beliefs about Y occurring, given X has occurred.
Probability Theory
Probability theory is a branch of mathematics concerned with analyzing random phenomena. The core of probability theory deals with quantifying how likely events are to happen, or how likely it is that a proposition is true. In terms of practical application, probability theory guides us in making predictions and strategic decisions under uncertainty.

This field is vast, covering concepts and tools such as random variables, expected values, probability distributions, and much more. It forms the theoretical underpinnings for statistics, which is its applied counterpart. In the context of our solved exercise, probability theory principles are at play when we explore and determine the likelihood of the events X and Y occurring together (their intersection).
Discrete Mathematics
Discrete mathematics encompasses a set of branches of mathematics that are used in the study of discrete objects. It is the backdrop for most computer science and information technology concepts because it includes topics like logic, set theory, and graph theory—puzzle pieces to understanding algorithms and data structures.

Probability is also a key area within discrete mathematics. It deals with discrete events, which are those that have distinct outcomes, such as the roll of a die (with six defined outcomes). The calculation of intersection probability, based on conditional probability as shown from the example of retrieving \(P(X \cap Y)\), is a practical demonstration of discrete mathematics at work. Students learn to rigorously prove and use these principles to effectively analyze real-world scenarios.

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

A person pays \(\$ 1\) to play the following game: The person tosses a fair coin four times. If no heads occur, the person pays an additional \(\$ 2\), if one head occurs, the person pays an additional \(\$ 1\), if two heads occur, the person just loses the initial dollar, if three heads occur, the person wins \(\$ 3\), and if four heads occur, the person wins \(\$ 4\). What is the person's expected gain or loss?

Let \(n=p_{1}^{k_{1}} p_{2}^{k_{2}} \cdots p_{m}^{k_{m}}\) where \(p_{1}, p_{2}, \ldots, p_{m}\) are distinct prime numbers and \(k_{1}, k_{2}, \ldots, k_{m}\) are positive integers. How many ways can \(n\) be written as a product of two positive integers that have no common factors a. assuming that order matters (i.e., \(8 \cdot 15\) and \(15 \cdot 8\) are regarded as different)? b. assuming that order does not matter (i.e., \(8 \cdot 15\) and \(15 \cdot 8\) are regarded as the same)?

Prove that if \(S\) is any sample space and \(U\) and \(V\) are events in \(S\) with \(U \subseteq V\), then \(P(U) \leq P(V)\).

Suppose that you placed the letters in Example 6.4.11 into positions in the following order: first the \(M\), then the \(I\) 's, then the \(S^{\prime}\) 's, and then the \(P\) 's. Show that you would obtain the same answer for the number of distinguishable orderings.

One urn contains 12 blue balls and 7 white balls, and a second um contains 8 blue balls and 19 white balls. An urn is selected at random, and a ball is chosen from the un. a. What is the probability that the chosen ball is blue? b. If the chosen ball is blue, what is the probability that it came from the first urn?

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.