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
ANSWERED
asked 2021-05-18

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 \(P(x<\frac{1}{3})\)

Expert Answers (1)

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 \(P(X<\frac{1}{3})\)
\(P(X<\frac{1}{3})=\int_{0}^{\frac{1}{3}}f(x,y)dx\)
\(=\frac{1}{3}\int_{0}^{\frac{1}{3}}(2x+3y)dx\)
\(=\frac{1}{3}[x^{2}+3xy]_{0}^{\frac{1}{3}}\)
\(=\frac{1}{3}[\frac{1}{9}+y]\)
\(=\frac{9y+1}{27}\)

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