Bunuel
If \(0<x<2\) and \(0<y<2\) on the \(xy\)-plane, what is the probability that \(x+y<1\)?
A. \(\frac{1}{2}\)
B. \(\frac{1}{4}\)
C. \(\frac{1}{6}\)
D. \(\frac{1}{8}\)
E. \(\frac{1}{10}\)
Given:\(0<x<2\) and \(0<y<2\).
We can represent x = 2, and y = 2 using straight lines as shown.
The region represented by \(0<x<2\) and \(0<y<2\) is the square represented by OABC.
The line \(x+y = 1\) has its x-intercept and y-intercept at (1,0) and (0,1) respectively. The line has a negative, slope and is shown in the figure.
The region below the line represents \(x+y<1\). The common overlap of the region represented by \(x+y<1\) and the region represented by \(0<x<2\) and \(0<y<2\) is the triangle represented by OMN.
Required Probability = \(\frac{\text{Favorable Region}}{\text{Total Region}}\)
Favorable Region = Area of \(\triangle OMN\) = \(\frac{1}{2} * 1 * 1 = \frac{1}{2}\)
Total Region = Area of square OABC = \(2 * 2 \) = 4
Required Probability = \(\frac{\frac{1}{2}}{4} = \frac{1}{8}\)
Option DIs there any other way of solving this question? Why are you taking the whole square region and the triangle region. I am not able to understand