Find the Probability That Two Uniform Variables Have Product Above One-Half

Let $X$ and $Y$ be independent random variables, each uniformly distributed over $(0,1)$. What is the probability that $X \cdot Y > \frac{1}{2}$?

Constraints and Clarifications

Work with the continuous uniform distribution and the stated independence. Describe the relevant region or conditional probability and provide an exact expression. At the boundary, the strict versus non-strict inequality makes no difference because that event has probability zero.

Hint — Locate the feasible part of the square: For a fixed value of $X$, identify which values of $Y$ can satisfy the product inequality and when that interval is nonempty.

What a Strong Answer Covers

• The uniform joint density on the unit square and the region where the product inequality holds.
• Correct integration limits instead of integrating over impossible values of $X$.
• An exact probability, a reasonable numerical interpretation, and a simple bound that checks the result.

Follow-up Questions

1. How does the expression change for $P(X \cdot Y > t)$ with $0 < t < 1$?
2. Why is multiplying $P(X > 1/2)$ and $P(Y > 1/2)$ not the answer to the original question?

Overview: Compute the probability that two independent uniform variables have product above one-half by conditioning or integrating over the unit square.