Question

A discrete probability distribution for a random variable \(X\) is given. Use the given distribution to find \((a)\) \(P(X \geq 2)\) and \((b) E(X)\). $$ \begin{array}{l|llll} x_{i} & 0 & 1 & 2 & 3 \\ \hline p_{i} & 0.80 & 0.10 & 0.05 & 0.05 \end{array} $$

Short answer

(a) P(X ≥ 2) = 0.10, (b) E(X) = 0.35.

Step-by-step solution

Understand the Probability Distribution

Let's begin by understanding the given discrete probability distribution for the random variable \(X\). The values \(x_i\) are \(0, 1, 2, 3\), with corresponding probabilities \(p_i = 0.80, 0.10, 0.05, 0.05\) respectively.

Calculate P(X ≥ 2)

To find \(P(X \geq 2)\), we sum the probabilities of all outcomes where \(X\) is greater than or equal to 2. This includes \(P(X = 2)\) and \(P(X = 3)\). Thus, \(P(X \geq 2) = P(X = 2) + P(X = 3) = 0.05 + 0.05 = 0.10.\)

Apply the Expected Value Formula

To find the expected value \(E(X)\), use the formula \(E(X) = \sum x_i \cdot p_i\). This involves multiplying each value \(x_i\) by its corresponding probability \(p_i\), and then summing the results.

Calculate Each Component for E(X)

Calculate each component: - \(0 \times 0.80 = 0\) - \(1 \times 0.10 = 0.10\) - \(2 \times 0.05 = 0.10\) - \(3 \times 0.05 = 0.15\).

Sum the Components to Find E(X)

Add the components obtained in the previous step to find \(E(X)\): \(0 + 0.10 + 0.10 + 0.15 = 0.35\). Therefore, the expected value \(E(X) = 0.35.\)

Key concepts

Expected Value

The expected value, often denoted by \(E(X)\), is a fundamental concept in probability and statistics. It represents the average or mean value you would expect in a large number of trials of an experiment involving random variables. Imagine this as the center point of a probability distribution, where everything balances out over time. Calculating expected value involves a straightforward formula: you multiply each possible outcome of the random variable by its probability of occurring, and then sum all these products together. In other words, it is a weighted average of all possible outcomes. Each outcome is weighted according to its likelihood or probability.In the exercise, the expected value was calculated for a random variable \(X\) with possible outcomes \(x_i\) and their corresponding probabilities \(p_i\). The formula used was: \[E(X) = \sum x_i \cdot p_i\]If you apply this step by step: for \(x_0 = 0\), its contribution is \(0 \times 0.80 = 0\); for \(x_1 = 1\), \(1 \times 0.10 = 0.10\); for \(x_2 = 2\), \(2 \times 0.05 = 0.10\); and for \(x_3 = 3\), \(3 \times 0.05 = 0.15\). Adding these up, you get the expected value \(E(X) = 0.35\). This means that, on average, you would expect \(X\) to have the value 0.35 over many trials.

Probability Distribution

A probability distribution is essentially a map of all possible outcomes of a random variable, each associated with a particular probability. It gives us a complete picture of how values are spread out and the likelihood of each occurring.In this exercise, we have a discrete probability distribution associated with the random variable \(X\). The term "discrete" means that \(X\) can only take certain specific values. Here, \(X\) can be 0, 1, 2, or 3. Each of these values has an associated probability, denoted by \(p_i\), which tells us how likely each outcome is.To better understand, let's think about the probabilities here:

• \(P(X = 0) = 0.80\): There's an 80% chance of \(X\) being 0.
• \(P(X = 1) = 0.10\): There's a 10% chance of \(X\) being 1.
• \(P(X = 2) = 0.05\): Only a 5% chance that \(X\) will be 2.
• \(P(X = 3) = 0.05\): Similarly, a 5% chance for \(X\) to be 3.

Each of these probabilities adds up to 1 or 100%, indicating they cover all possible scenarios for this discrete variable.

Random Variable

A random variable is a key concept in probability, serving as a bridge between real-world random processes and mathematical analysis. It assigns numerical values to the outcomes of random phenomena. Additionally, these variables help in quantifying these results and performing further calculations.Think of a random variable as a function mapping outcomes of a stochastic (random) process to numbers. The word "variable" is used because the outcome can differ each time the experiment is conducted. This variability is the crux of probabilistic analysis, allowing us to deal systematically with uncertainty.In the exercise, \(X\) is defined as a discrete random variable, meaning it takes on a countable number of distinct values: \(0, 1, 2, \text{and } 3\). The distribution given indicates how likely each outcome is - for instance, \(X = 0\) has a much higher probability (0.80) compared to the others.Random variables can be either discrete, like in this example, where specific, separate values are possible outcomes, or continuous, where values within a range are possible. Understanding random variables is critical for analyzing data, predicting results, and making informed decisions based on probabilistic models.