91Ó°ÊÓ

Consider a game based on the days of a 31 -day month. A day is chosen at randomsay, by spinning a spinner. The prize is a number of dollars equal to the sum of the digits in the date of the chosen day, For example, choosing the 31 st of the month pays \(\$ 3+\$ 1=\$ 4,\) as does choosing the fourth day of the month. (a) Set up the underlying sample space \(\Omega\) and its probability density, the value of which at \(\omega\) gives the reward associated with \(\omega\). (b) Define a random variable \(X(\omega)\) on \(\Omega\) with a value at \(\omega\) that gives the reward associated with \(\omega\). (c) Set up a sample space \(\Omega_{X}\) consisting of the elements in the range of \(X,\) and give the probability distribution \(p x\) on \(\Omega_{X}\) arising from \(X\). (d) Determine \(P(X=6)\). (e) Determine \(P(2 \leq X \leq 4)=P(\omega: 2 \leq X(\omega) \leq 4)\). (f) Determine \(P(X>10)=P(\omega: X(\omega)>10)\).

Step by step solution

01

Define Sample Space \(\Omega\) and Probability Density

The sample space \(\Omega\) contains each day of the month from 1 to 31. Each day \(\omega\) has an equal probability of being chosen, as each day is equally likely to be selected at random. Therefore, the probability density function \(P(\omega)\) for each day is \(\frac{1}{31}\). The value of the reward at each \(\omega\) is the sum of the digits of that day. For example, for the 21st day, the sum is \(2 + 1 = 3\).

02

Define the Random Variable \(X(\omega)\)

The random variable \(X(\omega)\) associates each day \(\omega\) with the reward, which is the sum of the digits of that day. So, \(X(1) = 1\), \(X(21) = 3\), \(X(31) = 4\), and so on. \(X(\omega)\) provides the reward or payout for picking day \(\omega\).

03

Set up Sample Space \(\Omega_X\) and Probability Distribution \(p_X\)

\(\Omega_X\) consists of all possible values of \(X(\omega)\). These are the distinct sums of digits that can occur from 1 to 10. Calculate the frequency of each sum and the probability associated with each sum by dividing by 31. Example sums are 1 (day 1), 2 (days 2, 11, 20), up to 4 (days 4, 13, 22, and 31), and so forth.

04

Calculate \(P(X=6)\)

Find out which days result in a digit sum of 6. They are: 15, 24, and 33. Hence, the probability \(P(X=6) = \frac{3}{31}\).

05

Calculate \(P(2 \leq X \leq 4)\)

To calculate \(P(2 \leq X \leq 4)\), find the days that give sums 2, 3, and 4. They are: 2, 3, 11, 12, 20, 21, 30, 4, 13, 22, 31. Therefore, \(P(2 \leq X \leq 4) = \frac{11}{31}\).

06

Determine \(P(X>10)\)

Examine if any sum from days 1 to 31 exceeds 10. Since the maximum sum is 4, for instance, 31 gives 4, the probability \(P(X>10) = 0\).

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Key Concepts

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

Understanding Random Variables

A random variable is a fundamental concept in probability theory that assigns numerical values to outcomes of a random process. In the context of a probability scenario like choosing a day at random in a month, we define a function that maps each outcome (day) to a value, which in our case is the sum of the digits of the day. Imagine each day of the month being represented as a number; then, the random variable takes this input and gives an output that is a specific reward.

For example, if you randomly pick the 21st day of a month, the random variable gives you 3 as the reward, because the sum of 2 and 1 equals 3. This transformation from a real-world occurrence to a numerical value is what makes random variables such a powerful tool in probability analysis.

Exploring Sample Space

The sample space is essentially the set of all possible outcomes for a particular experiment. For the game we are discussing, the sample space \(\Omega\) consists of every single day in a 31-day month. Each day is a unique outcome that can be observed.

This concept is crucial because it defines the boundaries of the probability problem we are dealing with. By listing all possible days, we create a comprehensive view of what can happen. This allows us to calculate probabilities with precision, knowing we’ve considered every feasible outcome.

Probability Distribution Simplified

A probability distribution gives us a way to understand how the probabilities of different outcomes are spread across a sample space. In essence, it tells us how likely each outcome is in the grand scheme of things.

In our exercise, the probability distribution arises from the random variable that assigns a probability to each potential sum of digits found in the days of the month. Each value that the random variable \(X(\omega)\) can take (such as 1, 2, 3, etc.) has a specific probability associated with it, which tells us how likely it is for that sum of digits (or reward) to occur when a day is chosen at random.

Decoding Probability Density Function

A probability density function (PDF) is a function that describes the likelihood of a random variable to take on a particular value. It offers a way to define how the probabilities are distributed over the possible values in a continuous random variable setting.

In our problem, however, since we are dealing with discrete random outcomes (days of the month), the concept of a probability density function applies slightly differently. Here, it describes the probability of each distinct value of the sum of digits that can be seen from random day selection. This helps calculate exact probabilities, like finding out how likely it is to pick a day where the sum of the digits results in a specific number, thus extracting meaningful information from random processes.