Answer:
Probability=2/5
Step-by-step explanation:
Let the line segment be AB of uniform length and x is distance
so
[tex]P(\frac{min(x,L-x)}{max(x,L-x)} )<1/4[/tex]
Case 1 min(x,L-x) then
[tex]=\frac{x}{L-x}\leq 1/4\\x<L/5[/tex]
Case 2 min(x,L-x)=L-x then
[tex]=\frac{L-x}{x}<1/4\\ x>\frac{4L}{5}\\[/tex]
So Probability
[tex]P=\int\limits^{L/5}_0 {f(x)} \, dx+\int\limits^{L}_{4L/5} {f(x)} \, dx\\ P=\int\limits^{L/5}_0 {\frac{1}{L} } \, dx+\int\limits^L_{4L/5} {\frac{1}{L} } \, dx\\ P=\frac{1}{L}(\frac{L}{5} -0+L-\frac{4L}{5} )\\ P=\frac{1}{L} (2L/5)\\P=\frac{2}{5}[/tex]
