If X and Y are two random variables having density function \(\rm f(x,y)=\left\{\begin{matrix}\frac{(6-x-y)}{8},0<x<2,2<y<4\\\ 0,\rm elsewhere\end{matrix}\right.\) then P(X + Y < 3) is equal to:
\(\frac{5}{24}\)
Setting up the Probability Integral
We are given the joint probability density function:
\(f(x,y)=\frac{6-x-y}{8}\) for \(0<x<2,\ 2<y<4\), and 0 elsewhere.
We need \(P(X+Y<3)\), the integral of f over the region where the domain meets \(x+y<3\).
Determining the Region
The line \(x+y=3\) meets \(y=2\) at \(x=1\), and meets \(x=0\) at \(y=3\). So the region inside the rectangle satisfying \(x+y<3\) is:
\(0<x<1,\ 2<y<3-x\).
Inner Integral (with respect to y)
\(\int_{2}^{3-x}(6-x-y)\,dy=\left[(6-x)y-\frac{y^2}{2}\right]_{2}^{3-x}\)
At \(y=3-x\): \((6-x)(3-x)-\frac{(3-x)^2}{2}=\frac{(3-x)(2(6-x)-(3-x))}{2}=\frac{(3-x)(9-x)}{2}\).
At \(y=2\): \(2(6-x)-2=10-2x\).
Difference: \(\frac{(3-x)(9-x)-2(10-2x)}{2}=\frac{x^2-12x+27-20+4x}{2}=\frac{x^2-8x+7}{2}\).
Outer Integral (with respect to x)
\(P(X+Y<3)=\frac{1}{8}\int_{0}^{1}\frac{x^2-8x+7}{2}\,dx=\frac{1}{16}\int_{0}^{1}(x^2-8x+7)\,dx\)
\(=\frac{1}{16}\left[\frac{x^3}{3}-4x^2+7x\right]_{0}^{1}=\frac{1}{16}\left(\frac{1}{3}-4+7\right)=\frac{1}{16}\cdot\frac{10}{3}=\frac{10}{48}=\frac{5}{24}\).
Conclusion
Therefore \(P(X+Y<3)=\frac{5}{24}\), which matches option 2.
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