At some point C of a telephone line AB of length L a break…
At some point $C$ of a telephone line $AB$ of length $L$ a break has occurred. Find the probability that the distance from $C$ to $A$ is not less than $l$.
Explanation of Method
With no other information, the break is equally likely at every point of the line: the distance $x=AC$ is uniformly distributed on $[0,L]$. In one dimension, geometric probability is length divided by length, so we measure the part of the line where $x\ge l$.
Step 1: The model
Put $A$ at $0$ and $B$ at $L$. The break point $C$ is at $x\in[0,L]$, and every point is equally likely, so for any segment $I\subseteq[0,L]$
\[ P(x\in I) = \frac{\text{length of } I}{L}. \]
Step 2: The favourable segment
"$AC$ not less than $l$" means $x\ge l$: the segment $[l,L]$ of length $L-l$ (for $0\le l\le L$).
Step 3: The probability
\[ \boxed{P(AC\ge l) = \frac{L-l}{L} = 1-\frac lL, \qquad 0\le l\le L.} \]
(If $l\gt L$ the event is impossible, $P=0$.)
Step 4: Example
$L=10$ km, $l=3$ km: $P = 1-0.3 = 0.7$. The repair crew starting from $A$ will have to travel at least $3$ km in $70\%$ of cases.
Result
| Answer | |
|---|---|
| Probability | $P(AC\ge l)=1-l/L$ for $0\le l\le L$ |