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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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Why do you think your method ought to work? Without knowing that, it is hard to answer what's wrong with your reasoning. In particular, where does the $1/3$ come from? If it should have been $1/2$, then what you're finding is the probability that all three $X_i$s are $>1/2$, but what you're looking for is just the probablity that at least one of them is. – Henning Makholm Nov 26 '12 at 12:05
Whoops should be 1/2, and yes, your phrasing shows me just what I got wrong. THanks! – Eliyahu Nov 26 '12 at 13:45

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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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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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.

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