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Can anyone help me with this problem, I can't figure out how to solve it...

The distribution of a random variable $X$ is:

$F(x)$ = \begin{cases} 1-e^{-2x}, & \text{if $x$ $\geq$ 0 } \\ 0, & \text{if $x$ < 0} \\ \end{cases}

Determine the density function. Determine the probability of the following events: $X$ > 2, −3 < $X$ ≤ 4.

Thanks!

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What have you done towards solving the problem? What do you know about the relationship between the distribution function and the density function? –  Dilip Sarwate Dec 10 '12 at 22:43
    
Hint: $\Pr(X\gt 2)=1-\Pr(X\le 2)=1-F_X(2)$. –  André Nicolas Dec 10 '12 at 23:00
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What's the problem?

$F(x)=\int_{-\infty}^{x} {f(x)dx}$, where f(x) is density function. By definition, $F(a)=P(X\leq a)$, so $P(X>2)=1-P(x\leq 2)=1-F(2)$ and $P(-3 < X \leq 4)=P(X \leq 4)-P(X \leq -3)$

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