# Given CDF of random variable X, Find P(X≤2), and P(1<X<3)?

cumulative distribution function of the random X is given by

$$F(x) = \begin{cases} 1-(1+x)e^{-x} & \text{for } x\gt 0 \\ 0 & \text{elsewhere} \end{cases}$$

1. Find $$P(X \leq2 )$$
2. Find $$P(1\lt X\lt 3)$$

Hi, i am not sure of how to solve this problem. Do i just have to integrate the values in given cdf equation?

• What was a cumulative distribution function again...? – Saucy O'Path Mar 24 at 19:59

$$F_X(x)\triangleq\Pr\{Xand $$\Pr\{a
A cumulative distribution function shows the probability of $$X$$ being less than a given value $$x$$, ie $$P(X. Therefore $$P(X<2) = 1-(1+2)e^{-2} = 1-3e^{-2}$$ and $$P(1 = $$P(X<3) - P(X<1) = 1-(1+3)e^{-3}-(1-(1+1)e^{-1}) = 2e^{-1}-4e^{-3}$$