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A discrete random variable is a type of random variable that takes on a countable number of distinct values. These values are often integers and arise in scenarios where outcomes can be counted, like the roll of a die or the choice of a card from a shuffled deck. For a clearer understanding, think of a discrete random variable like a function that, instead of assigning numbers to people, assigns probabilities to outcomes.

• Each outcome is a possible value that the discrete random variable can take.
• In the given problem, this discrete random variable, which we call X, can take values from the set \( S = \{1, 2, 3, \ldots, 10\} \).
• The corresponding probability assigned to each outcome \( k \) is given by the function \( p(k) = \frac{k}{N} \).

By understanding what a discrete random variable is, you can grasp how probabilities are mapped to these countable outcomes. This forms the stepping stone for analyzing more complex probability distributions.

The sum of natural numbers is a crucial concept when working with probability distributions involving integers. Natural numbers are simply the set of positive integers \( \{1, 2, 3, \ldots \} \). Their sum can be calculated using a specific formula.When you consider a set of natural numbers, like \( 1, 2, 3, \ldots, n \), the total can be computed using:\[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \]- For example, to find the sum of the first 10 natural numbers: - Use \( n = 10 \) in the formula, leading to: \[ \frac{10 \times 11}{2} = 55 \]- This result is essential because it helps ensure that the total probability across all outcomes adds up to 1, which is a requirement for any valid probability mass function.Understanding how to sum natural numbers sheds light on how individual probabilities sum up to form a complete, unified picture of outcomes in probability theory, aiding in the validation and calculation of probability distributions.

A probability distribution is like a blueprint that shows all the possible outcomes of a random variable and the likelihood of each occurring. For discrete random variables, this is often presented in the form of a probability mass function (PMF).In the context of our exercise:- We have a discrete random variable, X, with a probability mass function \( p(k) = \frac{k}{N} \) for \( k \in S = \{1, 2, 3, \ldots, 10\} \).- The key feature of a probability distribution is that all probabilities sum to 1.- To determine \( N \), the exercise ensures the sum of all assigned probabilities equals exactly 1 by solving for \( N \) in the equation: \[ \frac{1 + 2 + \ldots + 10}{N} = 1 \]The steps revealed \( N = 55 \), making \( p(k) = \frac{k}{55} \). This ensures every probability is accurate, allowing calculations like \( P(X < 8) \), where it sums probabilities for outcomes from 1 to 7, forming a cumulative understanding of the variable's behaviour.Through this understanding, one learns how probability distributions function as foundational elements in deciphering the likelihood of different events, enhancing statistical insights.