Question

# Two random variables X and Y with joint density function given by:f(x,y)=\begin{cases}\frac{1}{3}(2x+4y)& 0\leq x,\leq 1\\0 & elsewhere\end{cases}Find P(x<\frac{1}{3})

Random variables

Two random variables X and Y with joint density function given by:
$$f(x,y)=\begin{cases}\frac{1}{3}(2x+4y)& 0\leq x,\leq 1\\0 & elsewhere\end{cases}$$
Find $$P(x<\frac{1}{3})$$

2021-05-19

The joint density function of random variables X and Y is :
$$f(x,y)=\begin{cases}\frac{1}{3}(2x+4y)& 0\leq x,\leq 1\\0 & otherwise\end{cases}$$
We have to find :
We have to find the value of $$P(X<\frac{1}{3})$$
$$P(X<\frac{1}{3})=\int_{0}^{\frac{1}{3}}f(x,y)dx$$
$$=\frac{1}{3}\int_{0}^{\frac{1}{3}}(2x+3y)dx$$
$$=\frac{1}{3}[x^{2}+3xy]_{0}^{\frac{1}{3}}$$
$$=\frac{1}{3}[\frac{1}{9}+y]$$
$$=\frac{9y+1}{27}$$