Math Problem Statement

Let the probability distribution of X be given by X 0 1 2 3 4 . P(X=x) K 2K 3K 5K 2K .i) Find the value of K .ii) Find P(x≥2), P(x<2), P(x≤3)

Solution

Given the probability distribution for the random variable XX:

X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X = x) & K & 2K & 3K & 5K & 2K \\ \end{array}$$ ### (i) Find the value of $$K$$ The sum of the probabilities for all possible values of $$X$$ must equal 1. Therefore, we can write the equation: $$K + 2K + 3K + 5K + 2K = 1$$ Simplifying the left side: $$13K = 1$$ Now, solve for $$K$$: $$K = \frac{1}{13}$$ ### (ii) Find $$P(X \geq 2)$$, $$P(X < 2)$$, and $$P(X \leq 3)$$ 1. **$$P(X \geq 2)$$:** This is the probability that $$X$$ is 2, 3, or 4. So: $$P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)$$ Substituting the values: $$P(X \geq 2) = 3K + 5K + 2K = 10K$$ Using $$K = \frac{1}{13}$$: $$P(X \geq 2) = 10 \times \frac{1}{13} = \frac{10}{13}$$ 2. **$$P(X < 2)$$:** This is the probability that $$X$$ is 0 or 1. So: $$P(X < 2) = P(X = 0) + P(X = 1)$$ Substituting the values: $$P(X < 2) = K + 2K = 3K$$ Using $$K = \frac{1}{13}$$: $$P(X < 2) = 3 \times \frac{1}{13} = \frac{3}{13}$$ 3. **$$P(X \leq 3)$$:** This is the probability that $$X$$ is 0, 1, 2, or 3. So: $$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$$ Substituting the values: $$P(X \leq 3) = K + 2K + 3K + 5K = 11K$$ Using $$K = \frac{1}{13}$$: $$P(X \leq 3) = 11 \times \frac{1}{13} = \frac{11}{13}$$ ### Summary of Results: - $$K = \frac{1}{13}$$ - $$P(X \geq 2) = \frac{10}{13}$$ - $$P(X < 2) = \frac{3}{13}$$ - $$P(X \leq 3) = \frac{11}{13}$$ Would you like more details or have any questions? --- **Relative Questions:** 1. What is the probability $$P(X = 3)$$? 2. How do you interpret the cumulative distribution function (CDF) for this distribution? 3. What is the probability that $$X$$ is greater than 1 but less than 4? 4. Can you find the expected value $$E(X)$$ of this distribution? 5. How would the probabilities change if $$K = \frac{1}{10}$$? **Tip:** Always check that the sum of all probabilities equals 1 to ensure that the distribution is valid.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Probability Distributions

Formulas

Probability Mass Function (PMF)

Theorems

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Suitable Grade Level

High School

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