Question

Two people agree to meet on January 9, 2005 between 6:00 P.M. and 7:00 P.M., with the understanding that each will wait no longer than 20 minutes for the other. What is the probability that they will meet?

• A. \(\dfrac{5}{9}\)
• B. \(\dfrac{7}{9}\)
• C. \(\dfrac{2}{9}\)
• D. \(\dfrac{4}{9}\)

Hint:

Model both arrival times as points in a 60 by 60 square and find the area where the gap between them is at most 20 minutes.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept.
This is a geometric probability problem: since arrival times are continuous and random over an hour, we cannot count outcomes one by one. Instead we compare areas on a square grid of possible arrival-time pairs.

Step 2: Key Formula or Approach.
Let $x$ and $y$ be the minutes past 6:00 P.M. at which the two people show up, each uniform on $[0,60]$. Plot all possible pairs $(x,y)$ as points inside a $60 \times 60$ square, giving total area $3600$. They meet exactly when $|x-y|\le 20$, which is the band around the diagonal $x=y$. Instead of shading that band directly, it is simpler to first shade the two corner triangles where they miss each other, since those have a clean right-triangle shape.

Step 3: Detailed Explanation.
The miss each other region is $x-y>20$ or $y-x>20$.
For $x - y > 20$: the leftover corner beyond the line $x-y=20$ is a right triangle with legs of length $60-20=40$ each.
Area of this triangle $= \dfrac{1}{2} \times 40 \times 40 = 800$.
The mirror-image triangle for $y - x > 20$ has the same area, $800$, by the symmetry of the square.
Total miss area $= 800 + 800 = 1600$, so miss probability $= \dfrac{1600}{3600} = \dfrac{4}{9}$.
The meeting probability is everything else in the square:
\[ P(\text{meet}) = 1 - \frac{4}{9} = \frac{5}{9} \]

Step 4: Final Answer.
Working from the complement confirms the same result as working with the meeting band directly: the two people meet with probability $\dfrac{5}{9}$. \[ \boxed{\dfrac{5}{9}} \]