Let $X_1, X_2$, and $X_3$ be three independent, identically distributed random variables each with density function $f(x) = 3x^2$ for $0 \leq x \leq 1$. Let $Y = \max\{X_1,X_2,X_3\}$. Find $P[Y>1/2]$.
The real answer $511/512$. What's wrong with my method, below, that gives the wrong answer? $ P[Y>1/2] = P[X_1>1/2] \cdot P[X_2>1/2] \cdot P[X_3>1/2] $ because the random variables are independent, and $ P[X>1/2] = \int_{1/2}^1 3x^2\,\mathrm dx = 7/8 $ and $(7/8)^3 = 343/512$$.