Question

An exhibition was held in a hall on 15 August 2022 between 3 PM and 4 PM during which any person was allowed to enter only once. Visitors who entered before 3:40 PM exited the hall exactly after 20 minutes from their time of entry. Visitors who entered at or after 3:40 PM, exited exactly at 4 PM. The probability distribution of the arrival time of any visitor is uniform between 3 PM and 4 PM. Two persons \(X\) and \(Y\) entered the exhibition hall independent of each other. Which one of the following values is the probability that their visits to the exhibition overlapped with each other?

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

Hint:

Convert time windows to intervals and use geometric probability on the unit square.
When stay-time saturates at the end time, split cases by the threshold (here, 40 minutes).

Answer 1

The Correct Option is A

Step 1: Let the entry times (in minutes after 3 PM) be \(x,y \sim \text{Unif}[0,60]\). Each visitor stays on \([t,\,e(t)]\) where \[ e(t) = \begin{cases} t+20, & 0 \leq t < 40, \\ 60, & 40 \leq t \leq 60. \end{cases} \] Two visits do not overlap iff \(e(x)\le y\) or \(e(y)\le x\).

Step 2: For a fixed \(x \in [0,40)\): \(e(x) = x+20\). Non-overlap occurs for \(y \geq x+20\) or (if \(y<40\)) \(y \leq x-20\). Hence the measure in \(y\) is: \[ L(x) = \begin{cases} 40-x, & 0 \leq x < 20, \\ 20, & 20 \leq x < 40. \end{cases} \]

Step 3: For \(x \in [40,60)\): \(e(x)=60\). Non-overlap requires \(e(y)\leq x \;\Rightarrow\; y \leq x-20\) with \(y<40\). Thus: \[ L(x) = x-20, \quad 40 \leq x < 60. \]

Step 4: Compute the total non-overlap area in the \([0,60]\times[0,60]\) square: \[ A_{\text{no}} = \int_0^{20}(40-x)\,dx + \int_{20}^{40}20\,dx + \int_{40}^{60}(x-20)\,dx. \] Calculation: \[ A_{\text{no}} = 600 + 400 + 600 = 1600. \] Therefore: \[ \mathbb{P}(\text{no overlap}) = \frac{1600}{60^2} = \frac{4}{9}, \qquad \mathbb{P}(\text{overlap}) = 1 - \frac{4}{9} = \frac{5}{9}. \]

Final Answer: \[ \boxed{\; \mathbb{P}(\text{overlap}) = \tfrac{5}{9} \;} \]