Two ships must arrive at the same berth. The arrival times…

Two ships must arrive at the same berth. The arrival times of both ships are independent and equally likely at any moment of the day (24 hours). Find the probability that one of the ships will have to wait for the berth to become free, if the first ship stays at the berth for one hour and the second for two hours.

Explanation of Method

Let $x$ and $y$ be the arrival times (in hours) of the first and second ship; $(x,y)$ is uniform in the square $[0,24]^2$. If the first ship arrives first, the second must wait when it arrives less than $1$ hour later; if the second arrives first, the first must wait when it arrives less than $2$ hours later. The waiting region is a band around the diagonal, of different widths on the two sides. We subtract the two corner triangles where nobody waits.

Step 1: The model

Arrival times $x$ (first ship) and $y$ (second ship) are independent and uniform on $[0,24]$: $(x,y)$ is uniform in a square of area $24^2=576$.

Step 2: When does a ship wait?

First ship arrives first ($x\le y$): it occupies the berth until $x+1$. The second waits if $y\lt x+1$, i.e. $0\le y-x\lt1$.
Second ship arrives first ($y\le x$): it occupies the berth until $y+2$. The first waits if $x\lt y+2$, i.e. $0\le x-y\lt2$.

Waiting region: $-2\lt y-x\lt1$.

$24$ $24$ $x$ (first ship) $y$ $\tfrac12\cdot22^2$ $\tfrac12\cdot23^2$

one ship waits: $-2\lt y-x\lt1$

The square of arrival times (hours). The band around the diagonal is 1 hour wide above it and 2 hours wide below it; the white corner triangles have legs $23$ and $22$.

Step 3: Nobody waits

$y-x\ge1$: a triangle with legs $24-1=23$, area $\frac12\cdot23^2 = 264.5$.
$x-y\ge2$: a triangle with legs $24-2=22$, area $\frac12\cdot22^2 = 242$.

Step 4: The probability

\[ \boxed{P(\text{waiting}) = 1-\frac{\frac12(23^2+22^2)}{24^2} = 1-\frac{1013}{1152} = \frac{139}{1152}\approx0.121.} \]

Step 5: General formula

For a period of length $T$ and berth times $t_1$ (first ship) and $t_2$ (second ship):

\[ P = 1-\frac{(T-t_1)^2+(T-t_2)^2}{2T^2}. \]

With $t_1=t_2=t$ this becomes $1-(1-\frac tT)^2$, the meeting formula of Problem 87.

Result

	Answer
Waiting region	$-2\lt y-x\lt1$ in $[0,24]^2$
Probability	$P=139/1152\approx0.121$