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Let $X$ have density $$ f_X(x) = \begin{cases} \sqrt{3(x+2)}/6 & -2 \leq x \leq 1 \\ 0 & \text{otherwise}. \end{cases} $$ Find the probability that $X$ is positive.

Would this just be the integral from $0$ to $1$?

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    $\begingroup$ Yes- that would be the answer! $\endgroup$ – voldemort Apr 18 '14 at 4:59
  • $\begingroup$ @user139388 I do not know how to use HTML. Thanks for the edit. $\endgroup$ – Brent Apr 24 '14 at 0:55
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To find the probability that $X$ is positive implies that $\int^1_0 f_X(x)$ is what we are looking for.Evaluate that integral to get the probability that X is positive.

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Integrate the density function from 0 to 1, since it will not take any value for x > 1.
I think your answer will be $\displaystyle 1 - \frac{2 \sqrt{2}}{3\sqrt{2}} \approx 0.45567$

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