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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$$.

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    Whoops should be 1/2, and yes, your phrasing shows me just what I got wrong. THanks!2012-11-26

3 Answers 3

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It is not true that $ \{Y>1/2\}=\bigcap_{i=1}^3\{X_i>1/2\}. $ This were the case if $Y$ was the minimum. Instead you could look at $\{Y\leq 1/2\}$.

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I think you should rework $P[Y > 1/2]$ as $P[X_i > 1/2, X_j < X_i, X_k < X_i]$ (for $i,j,k$ all different) since what you want is the probability that the maximum is $>\frac{1}{2}$.

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If you rewrite it as $P(Y>1/2) = 1-P(Y \leq 1/2) = 1-P(X_1 \leq 1/2)P(X_2\leq1/2)P(X_3\leq1/2)$ Then you should be able to calculate the correct answer after deriving the CDF of X.