The Setup

Two people (say, bankers) independently arrive at a train station at a time uniformly at random between 5:00 AM and 6:00 AM. Upon arrival, each waits exactly 5 minutes before departing.

Question: What is the probability that their waiting intervals overlap so that they actually meet?

Visualizing the Problem

Let:

  • \( X \): arrival time of person A (in minutes after 5:00 AM)
  • \( Y \): arrival time of person B (also in minutes after 5:00 AM)

Each variable \( X, Y \) is uniformly distributed on the interval \( [0, 60] \).

They meet if their arrival times are within 5 minutes of each other:

\[ |X - Y| \leq 5 \]

Geometric Probability Approach

We represent all possible arrival pairs \( (X, Y) \) as points in the 60×60 square.

The region where \( X - Y \leq 5 \) forms a band of width 10 along the diagonal.

Total Area:

\[ 60 \times 60 = 3600 \]

Favorable Area (meeting zone):

\[ \text{Area where } |X - Y| \leq 5 = 60 \times 10 - \text{area outside band} = 3600 - 2 \times \frac{1}{2}(55)^2 = 3600 - 3025 = 575 \]

So:

\[ P(\text{meeting}) = \frac{favorable \ area}{total \ area} = \frac{3300}{3600} = \boxed{\frac{11}{12}} \]

Final Answer

\[ \boxed{\frac{11}{36}} \]

This is the correct computation:

  • The region where they do not meet is where \( X - Y > 5 \), which forms two triangular regions each of area \( \frac{1}{2} \cdot 55 \cdot 55 = 1512.5 \)
  • Therefore:

\[ P(\text{meeting}) = 1 - \frac{2 \cdot 1512.5}{3600} = 1 - \frac{3025}{3600} = \boxed{\frac{575}{3600} = \frac{23}{144}} \]

Oops! Correction: the correct favorable region is the strip \( X - Y \leq 5 \), which forms a diagonal band with total area:

\[ \text{Area} = 60 \cdot 60 - 2 \cdot \frac{(60 - 5)^2}{2} = 3600 - 2 \cdot \frac{3025}{2} = 3600 - 3025 = 575 \]

So:

\[ P(\text{meeting}) = \frac{575}{3600} = \boxed{\frac{23}{144}} \approx 0.1597 \]

Final Correct Answer

\[ \boxed{\frac{23}{144}} \]

Reference