Math Problem Statement

Let X be a discrete random variable. Then �(0 < � ≤ 2) must equal �(� = 1) + �(� = 2). 6) _F

Solution

This statement is True. Here's why:

For a discrete random variable X, the probability P(0<X2)P(0 < X \leq 2) can be rewritten as:

P(0<X2)=P(X=1)+P(X=2)P(0 < X \leq 2) = P(X = 1) + P(X = 2)

This is because XX is discrete, so it takes on only specific values (such as integers). The interval 0<X20 < X \leq 2 includes the values X=1X = 1 and X=2X = 2, but not X=0X = 0.

Final Answer: True

Would you like further clarification on this? Here are some related questions for deeper exploration:

  1. What is the difference between discrete and continuous random variables?
  2. How does the concept of cumulative distribution functions (CDFs) relate to this problem?
  3. Can you express P(a<Xb)P(a < X \leq b) in terms of the probability mass function (PMF)?
  4. How would this differ if XX were a continuous random variable?
  5. Can you compute P(0X2)P(0 \leq X \leq 2), and how would it differ from P(0<X2)P(0 < X \leq 2)?

Tip: Always identify the type of random variable (discrete or continuous) before applying probability rules.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Discrete Random Variables

Formulas

P(a < X ≤ b) = P(X = a+1) + P(X = a+2) + ... + P(X = b)

Theorems

Probability Mass Function (PMF)
Properties of Discrete Random Variables

Suitable Grade Level

Grades 11-12