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Let $X_1$, $X_2$, and $X_3$ be independent random variables with the continuous uniform distribution over $\left[0,2\right]$. Let $Z = \min\left(X_1, X_2, X_3\right)$.

Find $\mathbb{P}\left(Z \geq 0.5\right)$.

I had thought this was $\left(\frac{1}{4}\right)^3 = \frac{1}{64}$, but my answer was incorrect. Could anyone help me reach the correct answer with a short explanation?

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please use Latex –  Alex Apr 7 at 19:49
    
Hint: If the cdf of $X_i$ is denoted by $F(x)$, then the cdf of the minimum is given by $1−[1−F(x)]^n$. –  user130512 Apr 7 at 19:51
    
@Alex Don't know Latex, feel free to help out on the edit. Got a handy FAQ section or link with the latex formatting tips? –  CODe Apr 7 at 19:53
    
@CODe: did you try the 'help' button? –  Alex Apr 7 at 19:57
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@CODe: start with this: math.stackexchange.com/help/notation –  Alex Apr 7 at 20:02

1 Answer 1

up vote 1 down vote accepted

If $Z=\min X_n$, then $P(Z \geq r)=P(X_1 \geq r)\cdot P(X_2 \geq r) \ldots P(X_n \geq r)$. Do you see why?

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Ah, got it now. Thanks very much. –  CODe Apr 7 at 20:02

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