Question:
\(f_{x,y}(x,y)=\begin{cases} (x+y) & \quad 0\le x\le 1,0\le y\le 1 \\ 0 & \quad \text{otherwise } \end{cases}\)
The Probability \(P(x+y\le1)\) is:
\(f_{x,y}(x,y)=\begin{cases} (x+y) & \quad 0\le x\le 1,0\le y\le 1 \\ 0 & \quad \text{otherwise } \end{cases}\)
The Probability \(P(x+y\le1)\) is:
The Probability \(P(x+y\le1)\) is:
Updated On: Apr 28, 2025
- \(\frac{2}{3}\)
- \(\frac{1}{3}\)
- \(\frac{1}{6}\)
- \(\frac{1}{2}\)
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The Correct Option is B
Solution and Explanation
The correct option is(B): \(\frac{1}{3}\)
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