Question: A smart city integrator in San Francisco is analyzing traffic patterns and observes that 6 autonomous vehicles arrive at an intersection at random times during a 10-minute window. Each vehicle independently chooses a 2-minute interval (possibly overlapping) during the window: [0,10), and we are to compute the probability that all 6 vehicles arrive within some continuous 2-minute subinterval of [0,10), assuming uniform and independent arrival times.

["Title: Probability All 6 Autonomous Vehicles Arrive Within a 2-Minute Subinterval in San Francisco: A Smart City Insight", "---", "Introduction\nIn the evolving landscape of smart cities, understanding traffic flow is critical for optimizing urban mobility. A key challenge arises when analyzing arrival patterns of autonomous vehicles—especially in high-density urban centers like San Francisco. A recent analysis by a smart city integrator examines six autonomous vehicles independently choosing arrival times uniformly at random within the 10-minute interval [0,10). The central question is: What is the probability that all six arrive within some continuous 2-minute subinterval? This problem blends probability, uniform random variables, and geometric reasoning—offering a rich insight into urban traffic modeling.", "---", "The Problem Restated", "We model each of the 6 vehicles’ arrival times as independent uniform random variables on the interval $[0,10)$. The goal is to compute:", "$$\nP(\ ext{All 6 points lie within some interval of length } \leq 2 \ ext{ minutes})\n$$", "Unlike fixed-order arrival times, here we care only about the maximum and minimum arrival times:\nLet $X_{(1)} = \min(T_1, \ldots, T_6)$ and $X_{(6)} = \max(T_1, \ldots, T_6)$.\nThen, all six vehicles arrive within a 2-minute interval if and only if:", "$$\nX_{(6)} - X_{(1)} \leq 2\n$$", "Thus, the desired probability is:", "$$\nP(X_{(6)} - X_{(1)} \leq 2)\n$$", "---", "Geometric Probability Approach", "Because the arrival times are independent and uniform on $[0,10)$, we can treat them as six points sampled independently in a one-dimensional interval of length 10. The joint probability distribution is uniform over the 6-dimensional cube $[0,10)^6$.", "To compute $P(X_{(6)} - X_{(1)} \leq 2)$, we use geometric probability. The total volume of the sample space is $10^6$. We calculate the volume of the region where the range (difference between max and min) is at most 2.", "---", "Derivation via Integral Geometry", "Let $X_{(1)} = x$. Then, for all vehicles to nest within a 2-minute window ending at $x+2 \leq 10$, we require:", "- All $T_i \in [x, \min(x+2, 10))$\n- Since $x \geq 0$, we consider $x \in [0,8]$", "For fixed $x \in [0,8]$, the volume of allowable configurations within $[x, x+2)$ is $(2)^6 = 64$, because each vehicle independently chooses a time uniformly in a 2-minute interval.", "However, $x$ cannot be outside $[0,8]$, so we integrate over $x$ from 0 to 8:", "$$\n\ ext{Favorable volume} = \int_{x=0}^{8} \left(\ ext{length of } [x, \min(x+2, 10)) \right)^6 dx = \int_0^8 2^6 , dx = 8 \ imes 64 = 512\n$$", "This integral captures all configurations where a 2-minute interval starting at $x$ contains all arrivals. Note: this counts only intervals fully within $[0,10)$, avoiding edge overcounts.", "But since overlapping intervals are allowed and we only require existence of some such interval, this approach is valid due to the independence and uniformity—each configuration where arrivals fit in a 2-minute window is counted exactly once across all possible starting points, but our method historically works via normalized range distribution.", "An alternative and equivalent method uses known results from order statistics: for $n$ i.i.d. uniform $[0,L]$ variables, the cumulative distribution of the range $R = X_{(n)} - X_{(1)}$ is:", "$$\nP(R \leq r) = \sum_{k=1}^{\left\lfloor n/L \right\rfloor + 1} (-1)^{k-1} \binom{n}{k} \left(1 - \frac{k r}{L} \right)^{n-1} \quad \ ext{(for small } r\ ext{)}\n$$", "But for exact computation with $n=6$, $L=10$, $r=2$, we rely on geometric integration, which yields:", "$$\nP(R \leq 2) = \frac{ \int_0^{10} \left(\min(x+2,10) - x\right)^6 dx }{10^6} = \frac{512}{10^6} = 0.000512\n$$", "Wait—this underestimates because it assumes only intervals within $[0,10)$, but when $x \in [8,10)$, the interval $[x, x+2)$ extends beyond 10, violating bounds. Our geometric integration must restrict $x$ so that $[x, x+2] \subseteq [0,10)$, hence $x \in [0,8]$ as before.", "But the standard formula for the cumulative distribution of the range of $n$ uniform $[0,1]$ variables is:", "$$\nP(R \leq r) = \sum_{k=0}^{\left\lfloor n/(L - r) \right\rfloor} (-1)^k \binom{n}{k} \left(1 - \frac{k r}{L} \right)^{n-1}\n$$", "Scaling: Let $a = L = 10$, $b = r = 2$, $n = 6$. The probability that the range is at most $r$ is:", "$$\nP(R \leq r) = \sum_{k=0}^{\left\lfloor 6/(10 - 2) \right\rfloor = 0} \ ext{(simplified)} \quad \ ext{Not directly applicable.}\n$$", "Better: Use simulation or known tabulated values. For $n=6$, $L=10$, $r=2$, exact computation via order statistics gives:", "$$\nP(R \leq 2) = \frac{1}{10^6} \int_0^{10} \left( \min(x+2, 10) - \max(x, 0) \right)^6 dx\n$$", "Split at $x = 8$:", "- For $x \in [0,8]$: $\min(x+2,10) = x+2$, length = 2 → volume = $2^6 = 64$\n- For $x \in [8,10]$: $\min(x+2,10) = 10$, length = $10 - x$, so volume = $(10 - x)^6$", "Compute integral:", "$$\n\int_0^8 64 , dx + \int_8^{10} (10 - x)^6 dx = 8 \cdot 64 + \int_0^2 u^6 du = 512 + \left[\frac{u^7}{7}\right]0^2 = 512 + \frac{128}{7} \approx 512 + 18.2857 = 530.2857\n$$", "Wait—this contradicts earlier. Evaluation error:", "$$\n\int_8^{10} (10 - x)^6 dx = \int_0^2 u^6 du = \left[\frac{u^7}{7}\right]0^2 = \frac{128}{7} \approx 18.2857\n$$", "So total integral = $8 \cdot 64 + 18.2857 = 512 + 18.2857 = 530.2857$", "But total sample space is $10^6 = 1,000,000$, so:", "$$\nP = \frac{530.2857}{1,000,000} \approx 0.0005302857\n$$", "However, this counts all configurations where the maximum minus minimum ≤ 2, which is exactly $P(R \leq 2)$.", "Thus,", "$$\nP = \frac{1}{10^6} \left( 8 \cdot 64 + \int_8^{10} (10 - x)^6 dx \right) = \frac{512 + \frac{128}{7}}{10^6} = \frac{512 + 18.2857}{10^6} \approx 0.0005303\n$$", "But this cannot be correct—intuition says higher probability. Re-evaluate.", "Correct Normalization: The formula $P(R \leq r) = \sum = x$. For all arrivals to fit in $[x, x+2]$, we require $x \geq 0$, $x+2 \leq 10 \Rightarrow x \leq 8$, and each $T_i \in [x, x+2)$, so length 2.", "But due to overlap over different $x$, the proper method uses the }^{\left\lfloor n/L \right\rfloor} (-1)^k \binom{n}{k} \left(1 - \frac{k r}{L} \right)^{n-1}$ assumes continuous uniform and is derived from inclusion-exclusion over gaps. But for discrete sampling of interval positions, the correct continuous approximation with uniform spacing applies.", "Alternate proof: Fix the minimum $X{(1)probability that the range $R \leq 2$ for $n=6$, $L=10$:", "Standard result: For $n$ i.i.d. $U[0,L]$, the CDF of range is:", "$$\nP(R \leq r) = \frac{n}{L^n} \sum_{k=0}^{\left\lfloor n/L \right\rfloor} (-1)^k \binom{n}{k} (L - k r)^n \quad \ ext{for } kr \leq L\n$$", "But no—this is not standard."]