Question

Let \( X_1, X_2, X_3 \) be i.i.d. \( U(0,1) \) random variables. Then \[ P(X_1>X_2 + X_3) \]

• A. \( \frac{1}{6} \)
• B. \( \frac{1}{4} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{1}{2} \)

Hint:

When calculating probabilities for uniform random variables, break the problem into smaller parts and use integration for continuous random variables.

Answer 1

The Correct Option is A

Step 1: Define the probability.
We are asked to calculate \( P(X_1>X_2 + X_3) \) for three i.i.d. random variables, where \( X_1, X_2, X_3 \) are uniformly distributed on the interval [0,1]. Step 2: Set up the integral.
We can set up the probability as a double integral over the possible values of \( X_2 \) and \( X_3 \). For the uniform distribution: \[ P(X_1>X_2 + X_3) = \int_0^1 \int_0^{1-x_3} dx_2 dx_3 \] where the limits of integration reflect the fact that \( X_2 \) and \( X_3 \) are uniformly distributed and \( X_1 \) is greater than \( X_2 + X_3 \). Step 3: Solve the integral.
Solving this gives the result \( \frac{1}{3} \). Step 4: Conclusion.
Thus, the correct answer is \( \boxed{ \frac{1}{3} } \).