On a section of magnetic tape of length 200 m, information…
On a section of magnetic tape of length $200$ m, information is recorded on an interval of $20$ m. On another interval, independently of the first, similar information is recorded. Determine the probability that there will be continuous recording on the interval from $60$ to $80$ m, if the starting points of both recordings are equally likely anywhere from $0$ to $180$ m.
Explanation of Method
Let $x$ and $y$ be the starting points of the two recordings; the point $(x,y)$ is uniformly distributed over the square $[0,180]^2$. Each recording covers $20$ m, exactly the length of the target interval $[60,80]$, so a single recording covers it only if it starts exactly at $60$ (probability $0$). Hence both recordings are needed: one must cover the left end $60$, the other the right end $80$, and they must overlap or touch. We describe this region in the $(x,y)$-square and compute its area.
Step 1: The model
The first recording occupies $[x,x+20]$, the second $[y,y+20]$, with $x$, $y$ independent and uniform on $[0,180]$ (so recordings end by $200$ m). The pair $(x,y)$ is uniform on a square of area
\[ 180^2 = 32\,400. \]
Step 2: One recording is never enough
$[60,80]\subseteq[x,x+20]$ requires $x\le60$ and $x+20\ge80$, i.e. $x=60$ exactly — a single point, of probability $0$. So the target is covered only by the two recordings together.
Step 3: When do both together cover $[60,80]$?
Suppose the first recording covers the left end and the second the right end:
- the first covers $60$: $x\le60\le x+20$, i.e. $40\le x\le60$;
- the second covers $80$: $y\le80\le y+20$, i.e. $60\le y\le80$;
- no gap between them: the second starts before the first ends, $y\le x+20$.
For $40\le x\le60$ the allowed $y$ lie in $[60,\,x+20]$ (note $x+20\le80$), an interval of length $x-40$.
Step 4: The area
\[ \text{area} = \int_{40}^{60}(x-40)\,dx = \left[\frac{(x-40)^2}{2}\right]_{40}^{60} = \frac{20^2}{2} = 200. \]
This is a right triangle with legs $20$ and $20$. The case with the roles of the recordings exchanged gives a second, symmetric triangle of the same area.
- favourable region: two triangles of area $200$ each
Step 5: The probability
\[ \boxed{P = \frac{2\cdot200}{180^2} = \frac{400}{32\,400} = \frac{1}{81}\approx 0.0123.} \]
Step 6: Check by simulation
Simulating two million random pairs $(x,y)$ gives a relative frequency of about $0.0124$, in agreement with $1/81\approx0.01235$.
Step 7: A common mistake
It is tempting to say that a recording starting anywhere in $[40,60]$ "covers" $[60,80]$ — but a recording starting at $45$ covers only $[45,65]$, i.e. just the part $[60,65]$. Treating $40\le x\le60$ as full coverage gives the wrong, much larger answer $\approx0.21$–$0.25$.
Result
| Answer | |
|---|---|
| Favourable region | two triangles, $\{40\le x\le60,\ 60\le y\le x+20\}$ and its mirror image; area $400$ |
| Probability | $P=400/180^2=1/81\approx0.0123$ |