Conditional joint probabiltiy of a given pair The bivariate PDF of a random pair $(X, Y)$ is given by:
$f_{X,Y}(x,y) = 2e^{-x}e^{-2y}$ , $x\ge0, y\ge0$ What is the probability $Y < 4$ given  $X > 1$?
I calculated the conditional probability as $f_{Y\mid X}(y) = \frac{f_{X,Y}(x,y)}{f_X(x)}$
From using the above formula I got $f_{Y\mid X}(y) = 2e^{-2y}$, with $f_X(x) = e^{-x}$
I am confused on how to calculate the probability now that I have the equation $f_{Y, X}$
 A: The two answers posted here so far make the matter more complicated than it is.
The fact that the density can be factored as a product of a function of $x$ and a function of $y$ tells you immediately that $X$ and $Y$ are independent.
Therefore $\displaystyle \Pr(Y<4\mid X>1) = \Pr(Y<4) = \int_0^4 f_Y(y)\, dy.$
And since the density can be factored in that way, you get $f_Y(y) = (\text{constant} \cdot e^{-2y}).$ You can find the constant by using the fact that $\displaystyle \int_0^\infty f_Y(y)\, dy = 1.$
A: Well I am not an expert of probability but I think this could be a potential answer.
Given $f_{X,Y}(x,y)=2e^{-x}e^{-2y},~x\ge0,y\ge0$, we first compute the probability $P(Y<4 \cap X>1)$:
$$P(Y<4 \cap X>1)=P(Y<4\mid X>1)\cdot P(X>1)=\int_0^4 \int_1^\infty2e^{-x} e^{-2y} \, dx \, dy = \int_0^42e^{-2y}[-e^{-x}]_1^\infty dy= \int_0^42e^{-2y}e^{-1}\,dy=\frac{1}{e}[-e^{-2y}]_0^4=\frac{1}{e}(1-e^{-8})$$
Then we compute the marginal distribution of X
$$\int_0^\infty f_{X,Y}(x,y)\,dy = \int_0^\infty 2e^{-x}e^{-2y}\,dy=e^{-x}[-e^{-2y}]_0^\infty=e^{-x}$$
Then, 
$$P(X>1)=\int_1^\infty e^{-x}\,dx=[-e^{-x}]_1^\infty=0+e^{-1}=e^{-1}$$
Finally, we obtain the conditional probability 
$$P(Y<4\mid X>1) = \frac{P(Y<4 \cap X>1)}{P(X>1)}=\frac{\frac{1}{e}(1-e^{-8})}{\frac{1}{e}}=1-e^{-8}$$
Actually, it is notable that in this case, $X$ and $Y$ are independent but it is not in the scope of this question. Hope it helps.
A: From $f(x,y)$, we see that $X$ and $Y$ are independent. So \begin{align} \mathbb{P}(Y<4 \mid X>1) & =\mathbb{P}(Y<4) = \int_0^4 \mathrm{d}y \int_0^\infty \mathrm{d}x \;f(x,y) \\[10pt] & =\int_0^4 \mathrm{d}y \,2e^{-2y}\int_0^\infty \mathrm{d}x \;e^{-x}= 1 -e^{-8}.\end{align}
A: $\begin{split}\mathsf P(Y<4\mid X<1) &= \dfrac{\int_0^1\int_0^4 f_{X,Y}(x,y)~\mathsf d y~\mathsf d x}{\int_0^1\int_0^\infty f_{X,Y}(x,y)~\mathsf d y~\mathsf d x} 
\\[1ex] &= \dfrac{\int_0^1\int_0^4 2 e^{-x}e^{-2y}~\mathsf d y~\mathsf d x}{\int_0^1\int_0^\infty 2 e^{-x}e^{-2y}~\mathsf d y~\mathsf d x}
\\[1ex] &= \dfrac{\int_0^1 e^{-x}\mathrm d x\cdot\int_0^4 2e^{-2y}~\mathsf d y}{\int_0^1 e^{-x}\mathrm d x\cdot\int_0^\infty 2e^{-2y}~\mathsf d y}
\\[1ex] & = \int_0^4 2e^{-2y}~\mathsf d y
\\[1ex] & = 1-e^{-4}
\end{split}$
