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

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